Our Dynamic Universe Contents Section Equations of Motion
Equations of motion for objects with constant acceleration in a straight line. Motion-time graphs for motion with constant acceleration. Motion of objects with constant speed or constant acceleration.
Forces, energy and power
Balanced and unbalanced forces. The effects of friction. Terminal velocity. Resolving a force into two perpendicular components. Work done, potential energy, kinetic energy and power.
Collisions and explosions
Elastic and inelastic collisions. Explosions and Newton's Third Law. Impulse.
Gravitation
Projectiles and Satellites. Gravity and mass.
Special relativity
Introduction to special relativity.
The expanding Universe
The Doppler effect and redshift of galaxies. Hubble's law. Evidence for the expanding Universe.
Big Bang Theory
The temperature of stellar objects. Evidence for the Big Bang.
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Section 1 - Equations of motion Vectors The physical world can sometimes be expressed as a simple algebraic equation. For example V = IR. But the situation: “A bear walks 30 m in a straight line, pauses then walks 40 m in a straight line” tells us little about where the bear is now! He could be anywhere between 10 m and 70 m from his starting point and in what direction? 30 m
40 m
Many physical quantities are like this and can only be used mathematically if their direction is taken into account. They are referred to as vector quantities. For example: All forces (e.g. weight and friction), field strengths (gravitational, magnetic etc) and also, all aspects of motion. For motion we use the terms: symbol displacement velocity acceleration
s v a
unit _ m at (_ °) _ m s-1 at (_ °) _ m s-2 at (_ °)
“a passenger is jogging at 3 m/s on the deck of a ship travelling at 4 m/s” needs directions in each case to make the resultant velocity calculable. A physical quantity which has only magnitude but no direction is referred to as a scalar. Examples are: mass, electric charge, all forms of energy and pressure. “two boxes of apples, one of 3 kg and another of 4 kg” can only combine to equal 7 kg. Direction has no bearing. Special Scalars Regarding motion, we may wish to use measurements which exclude direction. Perhaps because our end point is of no interest, perhaps because the motion is along a defined path such as a road, perhaps we are only interested in the energy of an object and the direction it is travelling is irrelevant to us. We use these terms: symbol unit distance travelled m d instantaneous speed m s-1 v average speed m s-1 ¯v .Our Dynamic Universe
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Remarks on Motion average speed
d =t;
s average velocity = t Scalar Vector distance displacement speed velocity acceleration Vector Addition The rule for adding vectors can be derived from the way that displacements are added. The method is best described by an example. “A bear walks 30 m in a straight line towards a seal hole, then sees a better prey at 90° to its right and walks 40 m in a straight line towards it”. What is its displacement from its starting point? Scale Diagram On a sheet of paper (graph paper) draw a line 30 units (cm?) long.
Where the line ends, draw a 40 unit line at right angles to the right.
The resultant displacement is represented by a line from the starting point to the finishing point. Its magnitude is found from the length of the line and this line has the same direction as the resultant. Measure the length with a ruler and the direction using a protractor. Answer: displacement = 50 m at 53° to the right of the original direction. This method can be used for all vector addition. Trigonometry Roughly draw the shape of vector addition. Label the diagram and calculate the answer using trigonometry. This method may be slightly quicker. The increased precision is not usually useful unless the original vectors were measured extremely accurately.
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Resolving Vectors Just as two vectors can be added to produce a Resultant, a vector can be broken down into two imaginary components at right angles to each other. This process is called Resolving. This process is very important when dealing with Projectiles and Forces on a Slope. Resolving vectors makes use of trigonometry relationships.
F
F sin θ
θ F cos θ
Direction Conventions Where vectors are horizontal it is customary to quote a heading or bearing in degrees (to the nearest degree) clockwise from North. Always with 3 digits. e.g.: N quoted as: 40 N at (075°)
75° 40 N
Where vectors are vertical it is customary to quote an elevation either above or below the horizontal. quoted as:
40 m/s 25°
40 m/s at 25° above the horizontal
Summary of Vectors Scalar distance speed
Vector displacement velocity acceleration pressure force mass friction energy weight power momentum electric charge impulse temperature gravitational field strength .Our Dynamic Universe
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Motion in One Dimension We shall consider only motion in one dimension and importantly where acceleration is constant. Quantities are still vectors but direction is indicated only by a + or – sign. (on a diagram indicate the direction you have chosen to be positive by using
or
)
Equations of Motion A car accelerating from speed u to speed v over a time interval t has an acceleration a. 30
velocity (m/s)
v
25
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u
10
t
5
0 0
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a=
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tim e (s)
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v−u t
First Equation Rearranging this equation gives the first of the Equations of Motion:
v = u + at
Second Equation Using the idea of average velocity
v=
u+v s = 2 t
u+v s = 2 t
multiply by
v −u =a t
s u + v v − u =a t 2 t
simplify v2 − u2 = as 2
rearrange to give the second Equation of Motion v2 = u2 + 2as
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Third Equation From the first equation, substitute for v in the second equation: (u + at)2 = u2 + 2as
simplify u2 + 2uat + a2t2 = u2 + 2as
cancel u2 and divide by 2a ut + ½ at2 = s s = ut + ½ at2
Using the Equations of Motion Whenever we have an acceleration problem to solve we can see which of the 5 quantities are available to us by using the mnemonic suvat. We need to know 3 of these quantities to be successful and so can select the Equation of Motion which ignores the fourth. Example: A catapulted toy rocket accelerates at 8.4 m/s2 over a distance of 96 m and achieves a speed of 42 m/s. What was its initial speed? Translate the information into algebraic symbols using suvat: s
96
u
?
v
42
a 8.4 t
v = u + at
Select the Equation of Motion which contains the symbols we are interested in. v2 = u2 + 2as
v2 = u2 + 2as s = ut + ½ at2
422 = u2 + 2 × 8.4 × 96 u2 = 151.2 u = 12.29634 u = 12.3 m/s
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-100
displacement (m)
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displacement (m)
displacement (m)
Motion-time graphs Examples of velocity-time graphs with their associated displacement-time and acceleration-time graphs
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acceleration (m/s 2)
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The Bouncing Ball A ball will fall accelerating downwards at a constant 9.8 m/s2 due to its weight. When it reaches the ground it will experience a sharp upward force which accelerates it upwards until it loses contact with the ground.
500
500
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displacement (m)
displacement (m)
Remark on the difference in the two sets of graphs.
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0 0
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velocity (m/s)
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Section 2- Forces, Energy and Power Newton’ Laws Isaac Newton was able to encapsulate the concept of a Force in 3 statements. o
In the absence of a force constant velocity (no change in speed or direction)
o
Magnitude of a force F = ma (direction same as direction of acceleration)
o
By-product of a force reaction force (exactly opposite direction; exactly same magnitude)
From F = ma can be derived the definition of “one newton”. One newton is the force which causes a mass of one kilogram to accelerate at a rate of one metre per second squared. Free Body Diagrams A Free Body Diagram is a method of simplifying a situation involving several forces. It consists of a simplified diagram where objects with mass are reduced to a point or block. Three Examples:
1.
1.
Vertical rocket ship.
2.
Man in a lift
3.
Towing problem.
A rocket ship motor produces a thrust of 8700 N. It has a mass of 540 kg. What is its acceleration? thrust
8700 N or
weight
picture
540 kg 5292 N
Free Body Diagram Resultant F = 8700 – 5292 = 3408 F = ma 3408 = 540 a a = 6.3111 a = 6.31 m/s2
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2.
A man with mass 72 kg is standing on weighing scales in a lift. The scales show a force of 790 N. What is the acceleration of the lift?
790 N 72 kg 705.6 N
picture
Free Body Diagram
Resultant force = 790 – 705.6 = 84.4 F = ma a=
84.4 = 1.17222 72
a = 1.72 m/s2
3. A truck of mass 1320 kg tows a trailer of mass 850 kg and accelerates at 1.3 m/s2. The frictional force on the trailer is 375 N. What is the tension in the tow bar? 850 kg T
375 N 1.3 m/s2
picture
Free Body Diagram
F = ma = 850 × 1.3 = 1105 Resultant force = Tow force – trailer friction 1105 = T – 375 T = 730 N
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Forces in 2-Dimensions As forces are vectors. They obey the rules of vector addition but can be tricky to add because the forces on an object are visualised in a different way to that used in vector addition.
picture
vector addition diagram
Some people find it easier to work with forces by using the parallelogram method
Forces on a Slope There are two forces acting upon an object on a frictionless slope: weight and a force produced by the slope perpendicular to the slope. This perpendicular force must exactly balance the component of weight in the opposite direction as it doesn’t accelerate in that direction.
θ W
W
θ The only unbalanced force is the component of weight parallel to the slope: W sin θ Fparallel = W sin θ
Some prefer Fparallel = mg sin θ
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Example: A wooden block of mass 2.1 kg is placed on a slope at 30° to the horizontal. A frictional force of 4.6 N acts up the slope. The block slides down the slope for a distance of 3.2 m. Determine the speed of the block at the bottom of the slope. Ffriction
d
W
30°
Ffriction
Fun = Wparallel - Ffriction Wparallel
= mg sin θ - 4.6 Free Body Diagram
= 2.1 × 9.8 × sin 30° - 4.6 = 5.69 N
F 5.69 using F = ma a = m = = 2.7095 m/s2 2.1 using suvat s
3.2
u
0
v = u + at
v
?
v2 = u2 + 2as
v2 = u2 + 2as
s = ut + ½ at2
v2 = 02 + 2 × 2.7095 × 3.2
a 2.7095
v2 = 17.3410 t
v = 4.1642
v = 4.16 m/s
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Conservation of Energy The previous problem could have been tackled using conservation energy: A wooden block of mass 2.1 kg is placed on a slope at 30° to the horizontal. A frictional force of 4.6 N acts up the slope. The block slides down the slope for a distance of 3.2 m. Determine the speed of the block at the bottom of the slope. Energy change sliding down the slope Ep lost = work done against friction + Ek gained For which we need the height dropped.
h = d sin θ d
= 3.2 × sin 30° h
30°
Ep lost = work done against friction + Ek gained mgh = Ffrictiond + ½ mv2 2.1 × 9.8 × 1.6 = 4.6 × 3.2 + ½ × 2.1 × v2 v2 = 17.3410 v = 4.1642 v = 4.16 m/s
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Section 3 – Collisions and Explosions Momentum Momentum is a fundamental concept and, it turns out that by symmetry, it must be conserved. Momentum is calculated using p = mv (units are kg m s-1)
The Law of Conservation of Momentum In any interaction between objects total momentum of all objects before = total momentum of all objects after (in the absence of external forces)
Elastic and inelastic collisions In all collisions momentum is always conserved. In ordinary collisions kinetic energy is not conserved – some energy is converted to heat. Where no kinetic energy is lost, the collision is called “elastic” (typically this occurs only at sub-atomic level). Where kinetic energy is not conserved the collision is called “inelastic”.
Example: a)
A car of mass 1200 kg travelling at 11.2 m s-1 collides with a stationary car of mass 920 kg. If the cars lock together find their combined speed.
b)
By comparing the kinetic energy before and after the collision, find out if the collision is elastic or inelastic. Draw a simple sketch of the cars before and after the collision. BEFORE
AFTER
11.2 m/s
0 m/s
1200 kg
920 kg
v 1200 kg
920 kg
p = mv
p = mv
= 1200 × 11.2 + 0
= (1200 + 920) v
= 13440
= 2120 v by the Law of Conservation of Momentum total momentum before = total momentum after 13440= 2120 v v=
13440 2120
v = 6.3396 v = 6.34 m s-1
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b) Is the collision elastic or inelastic? BEFORE
AFTER
11.2 m/s
0 m/s
1200 kg
920 kg
-1
6.34 m s 1200 kg
920 kg
Ek =½ mv2
Ek =½ mv2
= ½ × 1200 × 11.22
= ½ × 2120 × 6.342
= 75264 J
= 42607 J
Kinetic energy is not the same, therefore the collision is inelastic
Momentum is a Vector We will only examine interactions in 1-dimension. Therefore direction is only positive or negative. By convention Right : Left
Example
Find the velocity following the collision. BEFORE
AFTER
5 m/s
4 m/s
20 kg
30 kg
v2 20 kg
30 kg
p = mv
p = mv
= 20 × 5 - 30 × 4
= (20 + 30) v2
= 100 – 120
= 50 v2
= -20 momentum before = momentum after -20 = 50 v2 v2 = −
20 50
v2 =-0.4 m/s (in other words 0.4 m/s to the LEFT)
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Impulse When an object is accelerated by a force F for a time t.
F = ma = m
v − u mv − mu = t t
Ft = mv − mu
applied force × time applied = change in momentum
Ft this property is called impulse
or change in momentum
unit of impulse N s or kg m s-1. The concept of impulse is useful in situations where a force is not constant and acts for a short period of time. One example of this is when a golf ball is hit by a club. During contact we might imagine that the force exerted by the club on the ball was constant. 12
force ( N )
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the value of Ft is the area under the graph. In practice the force on the ball varies with time
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force ( N )
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but the value of the impulse is still the area under the force-time graph
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average force
area = impulse time (s)
In simple impulse calculations an average force is used. The peak (maximum) force experienced is greater than this average value.
Examples 1.
In a game of snooker, the ball, of mass 0.22 kg, is accelerated from the rest to a velocity of 3.1 m s-1 by a force from the cue which lasts 54 ms. What is the size of the force exerted on the ball?
Ft = mv - mu F×
54 = 1000
0.22 × 3.1 - 0
F = 12.6296 F = 12.6 N
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2.
A tennis ball of mass 120 g, initially at rest, is hit by a racket. The racket is in contact with the ball for 35 ms and the force of contact varies over this period as shown in the graph. Determine the speed of the ball as it leaves the racket.
force(N)
430
35
time (ms)
Impulse = Area under graph = change in momentum 35
= ½ × 1000 × 430 = mv - mu 120
7.525 = 1000 × v - 0 v = 62.7083 v = 62.7 m s-1
3.
A fire hose delivers water at a rate of 43 kg/s at a speed of 52 m/s. Assuming it drops straight to the ground when it hits an obstacle what force does it exert on a door that it hits? Consider a time interval of one second. m = 4.3 kg t=1s Ft = ∆p F × 1 = 43 × 52 F = 2236 N F = 2240 N
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4.
Explain in terms of impulse why a cycle helmet may reduce brain damage! When a human skull is brought suddenly to rest a force acts upon the brain to change its momentum to zero. Without a helmet this occurs over a very short time interval. without helmet with helmet F1t1 = ∆p1
F2t2 = ∆p2 but ∆p1 = ∆p2 so F1t1 = F2t2
because the time to come to rest is much longer with a helmet the force will be much lower meaning less risk of brain damage
Summary
force(N)
impulse
∆p = mv – mu = change in momentum
area under graph
time (s)
Ft
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Section 4 – Gravitation The physics of projectiles
© Gustoimages/Science Photo Library A projectile is an object on which the only force acting is gravity. A projectile is any object, which, once projected, continues its motion by its own inertia and is influenced only by the downward force of gravity. By definition, a projectile has only one force acting on it: the force of gravity. Projectiles can be launched both horizontally and vertically and they have a combination of vertical and horizontal motions. Various experiments show that these horizontal and vertical motions are totally independent of each other. Closer study gives the following information about each component: Horizontal: constant speed Vertical: constant acceleration downwards (due to gravity) Let’s consider an example where the projectile starts with horizontal motion. Picture a motorcyclist on the top of a tall building about to perform a death-defying stunt of incredible skill. Do not try this at home! Predict her path as she leaves the roof. Explain your prediction.
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To summarise: Direction of motion
Forces
Velocity
Acceleration
Constant
None
Horizontal
Air resistance negligible so no forces in the horizontal
Changing with time
Vertical
Air resistance negligible so only force of gravity acting in the vertical
Constant or uniform acceleration of – 9.8 m s –2
Example: An object is released from an aircraft travelling horizontally at 1000 m s-1. The object takes 40 s to reach the ground. a) b) c) d)
What is the horizontal distance travelled by the object? What was the height of the aircraft when the object was released? Calculate the vertical velocity of the object just before impact. Find the resultant velocity of the object just before hitting the ground.
Before attempting the solution, you should divide your page into horizontal and vertical and enter appropriate information given or known.
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Another projectile situation to consider is the projectile at an angle to the horizontal.
© Gustoimages/Science Photo Library Remember that any vector can be resolved into its horizontal and vertical components:
The distance travelled horizontally (the range) is determined by using the cosine component of the launch velocity. The time of flight is determined by using the sine component of the launch velocity.
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To summarise, for a projectile at an angle to the horizontal: Direction of motion
Forces
Velocity
Acceleration
Constant
None
Horizontal
Air resistance negligible so no forces in the horizontal
Changing with time
Vertical
Air resistance negligible so only force of gravity acting in the vertical
Constant or uniform acceleration of – 9.8 m s –2
Projectiles and Newton’s thought experiment Think back to our death-defying motorcyclist. Consider what would happen if: • the building was taller • the horizontal launch velocity was greater • the Earth curved away more steeply.
This is what Newton (1643–1727) thought about. He considered that there is a horizontal launch velocity that would result in the projectile continually falling to Earth as a result of the force of gravity at the same rate at which the Earth’s surface curves away beneath it. The projectile would continually fall but would not hit the surface. It would therefore orbit the Earth. His thought experiment explained satellite motion around 300 years before it became a reality. On October 4 1957 Newton’s thought experiment became reality with the successful launch of the world’s first artificial satellite, Sputnik. Low-orbit and geostationary satellites are used for a wide range of applications, including environmental monitoring, communications, military applications and defence and scientific research.
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Understanding gravity Sir Isaac Newton (1642–1727) did not discover gravity. People had long been aware that objects when released would fall.
www.Cartoonstock.com Galileo (1564–1642) had carried out experiments on falling objects and put forward a theory on the motion of objects in freefall in his unfinished work De Motu (On Motion). Obviously, then, Galileo was performing experiments at the very beginning of his investigations into motion, and he took his experimental results seriously. Over the next two decades he changed his ideas and refined his experiments, and in the end he arrived at the law of falling bodies which states that in a vacuum all bodies, regardless of their weight, shape, or specific gravity, are uniformly accelerated in exactly the same way, and that the distance fallen is proportional to the square of the elapsed time. Newton’s significant contribution was to theorise that the force acting locally on an apple could be applied to the universe. Newton developed the universal theory of gravitation. This was published in his best-known work, the Principia. Questions to consider before you progress through this section: • • • • •
What is gravity? What is the force of gravity? What are the effects of gravity? What do we know about gravity? How can we make use of gravity?
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Mapping the gravitational field on Earth On 17 March 2009, the European Space Agency launched the Gravity Field and Steady State Ocean Circulation Explorer (GOCE), the lowest orbit research satellite in operation (at 254.9 km).
© ESA (see http://www.esa.int/esaLP/ESAYEK1VMOC_LPgoce_0.html). The mission of this satellite is to map the Earth’s gravitational field in greater detail than has previously been possible. This data will be used to: • inform predictions of climate • understand and monitor the effects of climate change, making accurate measurements of ocean circulation and sea level • develop an improved understand of the internal geology of the Earth, including hazards such as volcanoes • develop a system that allows comparison of heights all over the world to be made for the first time; such a system would be very useful when undertaking large-scale engineering projects such as bridge building and tunnel construction.
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In order to achieve its very challenging mission objectives, the satellite was designed to orbit at a very low altitude, where the gravitational variations are stronger closer to Earth. Since mid-September 2009, GOCE has been in its gravity-mapping orbit at a mere 254.9 km mean altitude – the lowest orbit sustained over a long period by any Earth observation satellite. To obtain clean gravity readings, there can be no disturbances from moving parts, so the entire satellite is a single extremely sensitive measuring device. ‘The gravity measuring system is functioning extremely well. The system is actively compensating for the effects of atmospheric drag and delivering a clean set of readings. The ‘standard’ acceleration due to gravity at the Earth’s surface is 9.8 m s –2 . The GOCE mission has established that the figure in fact varies from 9.788 m s –2 at the equator to 9.838 m s –2 at the poles. The Earth’s natural satellite: the Moon Greek philosophers understood that the Moon is a sphere in orbit around the Earth. They also realised that the Moon is not a light source, but reflects sunlight. Around 1850 years ago, Ptolemy (90–168) hypothesised that the Moon and Sun both orbit the Earth. This theory was the most commonly held belief in ancient Greece, Europe and in many other parts of the world. There is evidence that others, including Muslim scholars, developed an alternative theory, one in which the Earth was not orbited by the Sun, 250 years before Ptolemy proposed his theory. Evidence in support of the geocentric model Geocentrism is the belief that the Earth is the centre of the universe. What evidence supported this theory? What evidence do we now have to support an alternative theory? Certainly, the Sun, the Moon, the stars and other planets appear to revolve around us. In fact, we still commonly talk about the Sun in these terms: ‘rising’, ‘setting’, and ‘going down’. These suggest a belief that the Sun is orbiting the Earth. The Ancient Greeks also believed that the Earth is stationary. How do you know it isn’t? Evidence in support of a new model – heliocentrism With the work of Copernicus (1473–1543) the prevailing view of the universe began to change. Kepler (1571–1630) developed three laws that predicted that the orbits of the planets are elliptical, with the Sun at a focus of the ellipse. What evidence was there to support this model at the time when Copernicus and Kepler were developing these theories and laws? What evidence is there to support this model now? The Moon – what keeps it in place? The Moon remains in orbit around the Earth as a result of the force of gravity. It is the weakest of the four fundamental forces (the others being the strong force, the weak force and the electromagnetic force) yet it keeps the universe in shape! Newton developed the theory of universal gravitation. His theory hypothesised (and proved mathematically) that the Moon orbits the Earth as a result of the same force that causes an apple to fall from a tree. The moon falls around the Earth. .Our Dynamic Universe
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The law of universal gravitation Newton’s law of universal gravitation proposed that each body with mass will exert a force on each other body with mass. The theory states that the force of gravitational attraction is dependent on the masses of both objects and is inversely proportional to the square of the dstance that separates them. But Newton remained uncertain. He could not account for action at a distance without some medium, i.e. he was concerned about the distances over which he proposed gravitational force would act in space and the fact that space is a vacuum. F is force in newtons (N)
Gm1m2 F= r2
m 1 and m 2 are the two masses measured in kilograms (kg) r is the distance between them (m) G is the gravitational constant N m 2 kg –2
The value of the gravitational constant was determined by Cavendish (1731–1810) in the late 1700s. It was another hundred years before Boys (1855–1944) improved on its accuracy. G remains one of the most difficult constants to measure with accuracy. In 2007 a further value was published which suggested an improvement on the accuracy: G = 6.67 × 10 –11 N m 2 kg –2 This is the value we will use for calculations in Higher Physics. The force of gravity in everyday life The formula above allows us to calculate the force of gravity between point masses or spherical objects. Consider two objects, a folder of mass 0.3 kg and a pen of mass 0.05 kg, placed on a desk, 0.25 m apart. Calculate the magnitude of the gravitational force between the two masses (noting that the force is always attractive). Assume that the objects can be approximated to spherical objects.
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Gm1m2 F= r2
F is force in Newtons (N) = ? m 1 = 0.30 kg m 2 = 0.05 kg r = 0.25 m G = 6.67 × 10 –11 N m 2 kg –2
F=
6.67 × 10 –11 × 0.30 × 0.05 0.252
F = 1.60 × 10−11 N 6.67 × 10 –11 × 0.30 × 0.05 0.252 F = 1.60 × 10−11 N F=
This force is extremely small and thus has no effect on the objects. The gravitational force always acts in a straight line between the two objects being considered. Note that the gravitational force is always attractive, unlike electrostatic and magnetic forces. Consider the gravitational force due to the Earth acting on the pen. Assume that the Earth and pen can be approximated to spherical objects. F is force in newtons (N) = ?
Gm1m2 F= r2
m 1 = 5.97 × 10 24 kg (the mass of the Earth) m 2 = 0.05 kg r = 6.38 × 10 6 m (the radius of the Earth) G = 6.67 × 10 –11 N m 2 kg –2
6.67 × 10 –11 × 5.97 × 1024 × 0.05 (6.38 × 106 ) 2 F = 0.489N F=
How else could this calculation have been carried out? W = mg W = 0.05 × 9.8 W = 0.49 N
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The force of gravity on a subatomic scale F is force in newtons (N) = ?
Gm1m2 F= r2
m 1 = 1.67 × 10 –27 kg (mass of proton) m 2 = 1.67 × 10 –27 kg (mass of neutron) r = 0.84 × 10 –15 m (radius of proton; radius of neutron assumed equal) G = 6.67 × 10 –11 N m 2 kg –2 F = 2.64 × 10 –34 N
The force of gravity is clearly insignificant except where we are dealing with very large mass. At the subatomic scale, and at very short range, the strong force is the most significant. However, although weak, the gravitational force acts over enormous distances and on a universal scale! The force of gravity on a planetary scale The theory of universal gravitation can be used beyond satellite motion. It is commonly used in space travel in a technique called gravity assist. The gravity assist technique is often called, incorrectly, the ‘slingshot’ effect. The physics of the slingshot effect are considerably more complex. Gravity assist was made famous in the movie ‘Apollo 13’. The craft had been irreparably damaged by an explosion on board. The planned Moon landing was abandoned and the priority became the safe return of the astronauts to Earth. Gravity assist was the technique by which relatively small amounts of the limited available fuel could be used to manoeuvre the craft onto a trajectory that would allow the Moon’s gravitational field to turn the ship. Detailed information on the mission and the problems encountered is available from NASA. Gravity assist is routinely used to boost space flight on unmanned missions to distant planets such as Jupiter and Venus. To understand the principles of this, let us first consider a simple analogy. Consider a tennis ball travelling towards a tennis player holding her racquet at the ready. In this analogy, the ball is the spacecraft and the racquet is a massive planet. The racquet hits the ball. We will consider the outcome to be that the player successfully hits the ball, causing it to change direction and increase speed.
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Some questions (and answers): • What happens to the ball? (it changes direction and increases speed) • What type of interaction is this? (a collision) • What quantity is conserved in all collisions? (momentum) • What other quantity is considered in collisions? (energy) • In this case, what can we say about the momentum of the tennis ball? (it increases in magnitude) • And the momentum of the racquet? (the law of conservation of momentum tells us it must decrease in magnitude) • What about the energy of the tennis ball? (it increases) • What is the observable effect of the change in momentum of the tennis ball? (an observable change in speed. • Is there an observable effect of the change in momentum of the racquet? (no – because the racquet has a much larger mass than the ball) • Is there an effect of the change in momentum of the racquet? (yes – it slows down) In this case we have considered a mechanical interaction. The gravity-assist method is a gravitational exchange between the planet and the spacecraft. The spacecraft, which has very small mass, is able to gain momentum and energy from the planet, which has enormous mass. The mass difference is such that whilst the effect on the spacecraft (i.e. an increase in momentum and therefore speed) is observable and useable, the effect on the planet (i.e. a decrease in momentum and therefore speed) is not observable and in fact is negligible. Gravity assist makes use of the universal law of gravitation to enable space flight with minimal fuel requirements. This reduces the weight of the craft and allows us to reach greater distance than would otherwise be possible. The force of gravity on a universal scale We have explored the effects of the force of gravity on a small scale, its importance in satellite motion and its use in space flight. We have discussed some of the historical story associated with our understanding of gravitational force, but we have yet to discuss a very significant impact of gravitational force. Much of the current scientific thinking is that gravitational force is responsible for the formation of the solar system by aggregation (or accretion) of matter. Gravitational force also plays a crucial role in the birth and death of stars. The solar system formed around 4.5 billion years ago from a huge swirling cloud of dust. We know this because advances in technology, such as the Hubble telescope, have allowed us to look deep into space to observe the birth of stars similar to our sun.
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A huge cloud of dust Throughout the Milky Way, and other galaxies like it, are gigantic swirling clouds of dust and gas known as nebulae. It is within nebulae that stars are born. Our star, the Sun, was created in one such nebula. Something, perhaps the shock wave from an exploding supernova (dying star), triggered dust particles to be drawn together to form a dense spherical cloud. The accumulation of dust set off a chain reaction. As the core of the cloud attracted more dust, its gravitational pull increased. More and more dust was sucked in and the cloud collapsed in on itself. As this happened, the rotation of the cloud increased in speed, as happens when spinning ice-skaters pull in their arms. The rotational forces at the equator of the cloud prevented dust along this plane being drawn in, causing the cloud to flatten into a disc spinning around a dense core.
A pillar of dust and gas in the Orion Nebula. © NASA
A star is born As more and more mass is accumulated at the centre of the disc, the temperature increased dramatically. Eventually there was enough energy to set off nuclear reactions. Hydrogen atoms fused to form helium, releasing enormous amounts of energy in vigorous bursts. This marked the birth of the Sun, although it would take between 1 and 10 million more years for it to settle into the main sequence star recognisable today. The formation of the planets The planets, and other extraterrestrial objects such as asteroids, formed in the flat plane of the spinning disc of dust. Electrostatic forces or sticky carbon coatings made dust particles stick together to form clusters, which in turn stuck together to form rocks. Mutual gravity caused these rocks to come together, eventually to form planets. This ‘coming together’ of material is a process known as accretion (or aggregation).
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Section 5 - Special Relativity Einstein originally proposed his theory of special relativity in 1905 and it is often taken as the beginning of modern physics. It opened the door to whole new ways of thinking about the universe. This first theory of Einstein is termed ‘special’ because it only considers frames of reference moving at constant speed. His later general theory of relativity considers what happens if they are accelerating. Imagine you are on a train going at 60 mph: You are travelling at 60 mph when viewed by an observer at the side of the tracks, but you are at rest when viewed by the passenger sitting opposite you. These two points of view are known as frames of reference in physics, and relativity is all about how the physics measured in these frames of reference compares. If the observer was on a train travelling at 60 mph in the opposite direction then in their frame of reference you would be travelling at 120 mph. This works well for objects travelling at low speed but we find that it does not work so well for objects moving at speeds close to the speed of light. For instance, if two observers were passing each other in opposite directions at 2.0 × 10 8 m s –1 (from the reference frame of a stationary bystander) they would see the other travelling at 2.8 × 10 8 m s –1 rather than 4.0 × 10 8 m s –1 , as you might expect. History The idea of relativity has been around since 1632, when Galileo was trying to argue in favour of the rotation of the Earth around the sun. He considered the question of whether a scientist performing experiments in a sealed room on board a ship would notice any difference in the results if the ship was travelling at constant speed or at rest. Nowadays we might consider a train or aeroplane as the moving reference frame but it is worthwhile to note that the measurements we make in our laboratories are not in a universally stationary reference frame but on a rotating Earth (~950 km/h in Scotland) orbiting the Sun (~112,000 km/h), which is itself orbiting the centre of the galaxy (~800,000 km/h). When Newton first proposed his laws of motion in 1687, he adhered to Galileo's principle of relativity. However, this idea came under question in the 19th century, particularly when explaining how light travels. In 1864, James Clerk-Maxwell developed a theory of electromagnetism, which stated that light was a wave that travelled through some yet to be discovered universal medium called the aether (pronounced Eether). This aether could be considered a stationary master frame of reference that spread throughout the whole universe, including the vacuum of space. In 1881, JJ Thomson (who also discovered the electron) noticed that the mass of an object in motion increased. His work and Maxwell’s were developed further to show that the increase in mass depended on the speed of motion. Heaviside and Searle (1888–1897) also observed that a moving electromagnetic field contracted in the direction of travel. These experimental observations and others could be consistent with the widely accepted theory that included the idea of the aether. However, the aether itself was proving difficult to find and there was also much debate about the nature of the aether itself. Was it stationary or was it affected by the motion of objects through it, a bit like a boat dragging some water along with it?
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A very important experiment was performed by Michelson and Morley in 1887, where light was sent in two beams at right angles to each other to mirrors, so that they could recombine again at the centre. Because of the movement of the Earth around the Sun and its movement around the centre of the galaxy we would expect to see differences in the measurements as the Earth changed direction through the aether. However, this is now one of the most famous ‘failed’ experiments of physics as the two men were not able to prove the existence of the aether. Rather than accept that the aether didn’t actually exist, the scientists of their day considered that this was either down to lack of accuracy in the apparatus or that there were other theoretical reasons why the experiment didn’t work. However, there were a small number of scientists such as Lorentz and Poincaré who weren’t wholly convinced by the aether concept and were developing alternative theories. Einstein was also aware of Planck’s work in quantum physics, which stated that light could be thought of as a particle rather than a wave, an idea which did away with the need for an aether. This was also such a massive jump in thinking that there was a feeling that all the pillars of classical physics were open to challenge. So Einstein did not come up with his ideas in isolation but was widely influenced by other thinkers of his time. He was not an established scientist at this time but his work as a patent clerk involved a lot of study into synchronising clocks throughout the world, so he was often dealing with the relative nature of measurements of time. Although he had no access to scientific equipment he would consider the logical consequences of scientific assumptions. He called these his ‘thought experiments’ or Gedanken in German. The principles 1. 2.
When two observers are moving at constant speeds relative to one another, they will observe the same laws of physics. The speed of light (in a vacuum) is the same for all observers. This means that no matter how fast you go, you can never catch up with a beam of light, since it always travels at 3.0 × 10 8 m s –1 relative to you.
Einstein started with just these two principles. These were two almost philosophical statements, which he could not prove to be true, but using them he was able to derive his theory of special relativity.
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Time dilation A very simple thought experiment shows that one consequence of the speed of light being the same for all observers is that time experienced by all observers is not necessarily the same. There is no universal clock that we can all refer to – we can simply make measurements of time as we experience it. Imagine you are on a spaceship travelling at constant speed ( v ) relative to a colleague on a space station. You are investigating a timing device based on the fact that the speed of light is constant. You fire a beam of light at a sensor and time how long it takes to arrive. From your point of view the light has less distance ( l ) to travel than from the point of view of your colleague on the space station ( d ). If you both observe the speed of light as the same then you cannot agree on when it arrives, ie both of you experience time in a different way. movement of spaceship
sensor
l
sensor
d
light source
light source
Your point of view, where the source and sensor are moving along together
Your stationary colleague’s point of view, where the sensor moves after the light is fired
The time for the experiment as observed by your stationary colleague (t') is greater than the time observed by you when moving with the sensor (t), ie what you might observe as taking 1 second could appear to take 2 seconds to your stationary colleague. Note that you would be unaware of any difference until you were able to meet up with your colleague again. Your body processes would continue at the same rate. The formula linking these can be shown to be:
t' =
t
t = time, s 2
v 1− 2 c
t’ = observed time, s v = velocity of the object, ms -1 c = velocity of light, ms -1
γ =
1 v2 1− 2 c
Note this is often written as
t ' = tγ .Our Dynamic Universe
where
γ is known as the Lorentz Factor. 34.
Worked example A rocket is travelling at a constant 2.7 × 10 8 m s –1 compared to an observer on Earth. The pilot measures the journey as taking 240 minutes. How long did the journey take when measured from Earth? Solution: t = 240 minutes
t' =
v = 2.7 × 10 8 m s –1
t 1−
c = 3 × 10 8 m s –1
t' =
t' = ?
1−
v2 c2 240 ( 2 ⋅ 7 × 10 8 ) 2 (3 × 10 8 ) 2
t ' = 550 minutes
An observer on Earth would measure the journey as taking 550 minutes, ie 550 minutes would have passed from their point of reference. 2
v2 v Note: 2 is the same as , which can make the calculation simpler, particularly if the speed c c is given in the form of a fraction of the speed of light, eg the speed in the above example could 9
have been given as 0.9 times or 10 of the speed of light. The calculation can then be written as: 240 240 t'= = = 550 minutes 2 1 − 0 ⋅ 81 1− 0 ⋅9 Why do we not notice these time differences in everyday life? A graph of the Lorentz factor versus speed (measured as a multiple of the speed of light) is shown below.
We can see that for small speeds (ie less than 0.1 times the speed of light) the Lorentz factor is approximately 1 and relativistic effects are negligibly small. Even 0.1 times the speed of light is 300,000 m s –1 or 1,080,000 km h –1 or about 675,000 mph – a tremendously fast speed compared to everyday life. .Our Dynamic Universe
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However, the speed of satellites is fast enough that even these small changes will add up over time and affect the synchronisation of global positioning systems (GPS) and television broadcasters with users on the Earth. They have to be specially programmed to adapt for the effects of special relativity. Very precise measurements of these small changes in time have been performed on fast-flying aircraft and agree with predicted results within experimental error. Further evidence in support of special relativity comes from the field of particle physics, in the form of the detection of a particle called a muon at the surface of the Earth. These are produced in the upper layers of the atmosphere by cosmic rays (high-energy protons from space). Their measured lifetime is about 2·2 µ s and their speed is 99·9653% of the speed of light. Multiplying these together tells us that they should be able to travel about 600 m in their lifetime. However, muons are produced around 25 km above the surface of the Earth so they should not be detected at the surface. The fact that they are can be explained using time dilation. In the frame of reference of the muons, they survive for about 2·2 µ s, but from our perspective this time is 84 µ s because of to time dilation. This gives muons enough time to travel to the surface of the Earth. Length contraction Another implication of Einstein’s theory is the shortening of length when an object is moving. Consider the muons discussed above. Their large speed means they experience a longer lifetime due to time dilation. An equivalent way of thinking about this is that the muons experience the height of the atmosphere as smaller (or contracted) by the same amount as the time has increased (or dilated). A symmetrical formula for time dilation can be derived. Note that the contraction only takes place in the direction that the object is travelling:
v2 l' = l 1 − 2 c
l = length measured by moving object, m l’ = length measured by observer, m v = velocity of object, ms -1 c = velocity of light, ms -1
Ladder paradox There is an apparent paradox thrown up by special relativity: consider a ladder that is just longer than a garage. If we fly the ladder at high speed through the garage does length contraction mean that from our stationary perspective it fits inside the garage? How can this be reconciled with the fact that from the ladder’s reference frame the garage appears even shorter as it moves towards the ladder? The key to this question is simultaneity, ie whether different reference frames can agree on the time of particular events. In order for the ladder to fit in the garage the front of the ladder must be inside at the same time as the back of the ladder. Because of time dilation, the stationary observer (you) and a moving observer on the ladder cannot agree on when the front of the ladder reaches the far end of the garage or the rear of the ladder reaches the front. If you work out the equations carefully then you can show that even when the ladder is contracted, the front of the ladder and the back of the ladder will not both be inside the garage at the same time!
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Section 6 - The expanding universe ‘Why is the night sky dark?’ This question can be traced back to around 1576 and Thomas Digges, but it was first stated formally by the German astronomer Heinrich Olbers in 1823, hence the name given to this concept is Olbers’ Paradox. Prior to the expansion of the universe being demonstrated by Hubble in 1929, it was commonly assumed that the universe was:
infinite
eternal
static.
If this were true, no matter which direction you looked, your line of sight would eventually intersect with a star. The entire sky would be virtually as bright as the Sun! (Whilst the brightness of each star decreases with distance, this would be exactly compensated for by the increasing number of stars in any field of view.) The paradox cannot be resolved by arguing interstellar dust blocks out the light either, as Olbers himself did, since the dust would heat up and re-radiate the energy as infra red, leading to a much higher background radiation than that observed. It suggests that the universe is either: NOT infinite, NOT eternal or NOT static. This is the first of the observations that Big Bang theory seeks to explain.
The Doppler effect and redshift of galaxies The Doppler effect The Doppler effect is the apparent change in frequency of a wave when the source and observer are moving relative to each other. The effect is produced with all wave motions, including electromagnetic waves. It is probably most familiar in the context of sound and in particular the apparent change in frequency of a siren as an emergency vehicle approaches then passes. Or a Formula 1 racing car gives that characteristic eeeeee-owwwww as it approaches then passes. The effect is noticeable with any fast-moving vehicle, for example a train passing through a station or an aeroplane flying past at low altitude. Equations A number of equations can be derived for the Doppler effect. For a moving source there is an equation for the source moving towards the observer and another for the source moving away from the observer. Similarly for a moving observer there are two equations, one for away from and one for towards the source. For this course, the only equations required are for a stationary observer with the source moving away or moving towards the observer. These could be reduced to a single velocity equation (rather than two speed equations).
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Derivation: Stationary observer and source moving away Consider a loudspeaker emitting a constant tone:λ
stationary source
λ
moving source d
= observer = source λ = wavelength of source , f = observed frequency etc. and v = speed of sound Wavelength of source (using v=f λ )
v
λ = f ……………….[1]
When the source starts moving away, in the time between creating the first and the second wave (the period T), the source will have moved away from the observer by a distance: v 1 d = vT = f (using d = vt, then T = f ) v ie the wavelength increases by f so for the observer () v λ = λ + f Substituting for λ from [1] gives: v v λ = f + f (v + v) λ = f ……….…..[2] As the speed of sound is constant: v = fλ v λ = f Substituting for λ in [2] v (v + v) f = f rearranging: v f = f (v + v) .Our Dynamic Universe
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Summary For a source moving away from a staionary observer
v f observed = fsource (v + v
source)
where v is the speed of the waves, eg the speed of sound. By similar reasoning, for the source moving towards the observer:
v f observed = fsource (v - v
source)
☼ Practice Question Calculate the frequency of sound heard by a stationary observer when an ambulance siren emitting sound of frequency 840 Hz moves away from them at 13.4 ms-1(speed of sound 340 ms-1). v f observed = fsource (v + v
source)
340
= 840 × (340 + 13.4) = 808.149 = 808 Hz ☼ Practice Question Calculate the frequency of sound emitted by a fire engine which is travelling towards a stationary observer at 22.4 ms-1 if the observer hears sound of frequency 670 Hz. v f observed = fsource (v - v
source)
fsource = 670 ×
317.6 340
= 625.858 = 626 Hz The first Question gives the expected reduction in frequency observed and the second an increase in frequency observed which we associate with the change of pitch of a passing vehicle. If you are unsure whether to use the ‘+’ or ‘–’, check that their answer gives the expected increase or decrease (eeeeee-owwwww). We can also see from these equations that if the magnitude of v source is very small compared to v there is little effect on f observed . This is why for the effect to be noticeable, v source needs to be reasonably large in comparison with the speed of the waves (v).
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The Doppler Effect and electromagnetic waves The Doppler Effect also applies to electromagnetic waves, but at high speeds relativistic effects have to be taken into account. For non-relativistic speeds the equation reduces to the same as that for other types of waves. The redshift of a galaxy Line spectra Each element produces its own characteristic line emission and absorption spectra (this is covered in the Particles and Waves topic). for example: Helium
wavelength nm RED
ORANGE YELLOW
GREEN
BLUE
VIOLET
Redshift When studying distant celestial bodies, their spectra are found to contain the characteristic absorption lines of elements. However, sometimes the spectral lines are found to have been moved or shifted towards the red end of the spectrum because of an increase in wavelength (decrease in frequency).
increasing frequency
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This effect is called redshift and, if it is due to the Doppler effect, implies these bodies are moving away from us. The amount of the shift also allows us to calculate how fast they are moving away. Redshift is defined as the change in wavelength divided by the original wavelength, and given the symbol z. 1 2 3 4 blueshifted
redshifted 1 2 3 4
direction of movement of source Figure 1.7 Doppler Effect and redshift
∆ λ λ observed - λ source z= λ = λ source
So, redshift:
Redshift is a dimensionless quantity since it is a ratio of two lengths. Note: If there is a decrease in wavelength, ie the line spectrum has moved towards the blue end of the spectrum, this makes z negative, which means the body is moving towards us. This is referred to as a blueshift. For bodies travelling at non-relativistic speeds (less than about 10% of the speed of light) we can apply the Doppler equation for a stationery observer and a source moving away. We use a special symbol for the speed of light: c Using the Doppler equation we get: c f observed = f source c + v source f source c + vsource f observed = c
Rearranging
f source vsource f observed = 1 + c …………………..[1] Using c = f λ , rearranging
f=
f source f observed .Our Dynamic Universe
c
λ
=
the frequency ratio leads to:
λ observed …………………..[2] λ source 41.
Combining [1] and [2]
λ observed vsource =1+ c λ source λ observed vsource - 1= c λ source
or
λ observed - λ source vsource = c λ source But the definition of redshift is
z=
λ observed - λ source λ source
So
z=
vsource c
Note: This equation only applies to non-relativistic speeds, less than 10% of the speed of light. Radial speed The redshift can only tell us the radial velocity of the object.
Earth
radial velocity
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it tells us nothing about its tangential velocity
42.
Hubble’s law Measuring distance Units of distance Units of distance used in astronomy include: ☼ astronomical unit (AU) – 1 astronomical unit (AU) is the mean radius of the Earth’s orbit around the Sun. 1 AU is equivalent to 1.496 × 1011m ☼ light year (ly) – 1 light-year (ly) is the distance travelled by light in one year. One light year is equivalent to 9.46 × 1015m. The cosmic distance ladder Measuring distance to celestial objects is done by different methods, depending on how close the objects are. It is generally more straightforward to measure distance to near objects than more distant ones. It becomes increasingly difficult to measure distances the further away objects are so these require a different technique. It is a multi-step process, relying on measurements made on closer objects, hence the term ‘cosmic distance ladder’. ☼ radar or laser reflection - nearby planets ☼ parallax – nearby stars ☼ standard candle – further away Radar or laser This method is only suitable for distances up to a few AU. By sending a pulse of radio waves (radar or laser light) to an object and measuring the time it takes the pulse to reach the object and return, the distance can be calculated. Using distance d = vt where v is the speed of light and t is half the time for the pulse to reach the object and return; the distance to the object can be found. (Alternatively, use the full time to reach the object and return, then halve the answer for distance.) Parallax Close one eye and point at a small object several meters away. Without moving your head, open the other eye and close the first. Your finger is no longer pointing at the object. This effect is called parallax. Parallax is the apparent displacement of an object when viewed from different positions.
Earth planet θ
θ − parallax
A baseline on the Earth can be used and the geometry of the triangle used to calculate the distance to a nearby planet. .Our Dynamic Universe
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For measuring the distance to a nearby star, parallax can be used by taking sightings six months apart, so the baseline is the diameter of the Earth’s orbit round the Sun. The maximum angular displacement of the star from its mean position (when the angle between the Earth, Sun and the star is 90°) is called the annual parallax.
Earth (March)
DEarth’s orbit
Star Sun θ
Earth (Sep) θ − annual parallax
Earth’s orbit baseline
Parsec Angular measure:
1 60
of a degree is called 1 arcminute.
1 60
of an arcminute is 1 arcsecond.
A unit of distance which relates to parallax is the parsec (short for parallax-arcsecond). 1 parsec (pc) is the distance at which a star would have an annual parallax of 1 arcsecond. 1 parsec is equivalent to 3.09 × 10 16 m or 3.26 ly. The current limit of measuring distances with parallax is about 1000 pc, beyond which the parallax becomes too small to measure.
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Standard candles Luminosity is the total power emitted by a star, rather like the power rating of a light bulb. A standard candle is an astronomical object whose luminosity is known (a star or a supernova). With increasing distance an object appears less bright. Comparing the brightness of the standard candle with the brightness of a remote object allows a distance measurement to be made. Temperature–luminosity correlation The surface temperature of a star can be measured from the peak wavelength of its spectrum (see Section 7 - Big Bang theory). There is a high correlation between the absolute luminosity of a star and its surface temperature. Period–luminosity correlation One class of standard candle is the Cepheid variable star. These stars have a predictable variation of luminosity with time.
The period–luminosity correlation technique is similar to the temperature–luminosity correlation. Two main problems with using the standard candle are: ☼ the assumption of known luminosity of the standard candle ☼ absorption of light by intervening dust
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Hubble Edwin Hubble was an American astronomer who showed that there were galaxies beyond our own (these had previously been observed as nebulus objects but assumed to be placed between the nearby stars). He measured the distance to these galaxies and their recessional speed, based on their redshift. He found that most galaxies are receding from us and concluded that the universe is expanding.
The original graph published by Hubble in 1929 (white points for galaxies: black for the mean of presumed clusters) From his results, he also concluded that more distant galaxies are receding faster than closer ones. In fact he found that the recession speed (v) varied directly with distance (d), ie: v ∝ d, so v = H o d where H o is the Hubble constant. (It is the constant of proportionality of these data but it is presumed that its value is not constant outside this range). The discovery that the universe is expanding completely revolutionised twentieth century cosmology and provided the first piece of evidence for the Big Bang. 1.3.3 The age of the universe From Hubble’s rule, we can obtain an estimate of how long it took for a galaxy to reach its current position by assuming it has been moving away from us at its current speed constantly. d t=v Substituting v = H o d , we get: 1 t=H o 13.8 × 10 9 years which is consistent with more rigorous assessments of the age of the universe .Our Dynamic Universe
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Evidence for the expanding universe Measurements of the velocities Galaxies are moving away from us and the greater the distance from us the faster they are moving away. Assuming we are not in a privileged position (the centre of the universe) this is a clear indication that the universe is expanding. Space itself is expanding. One way of trying to imagine what is happening is to think of a balloon with spots on its surface. Blowing up the balloon causes each spot to move further apart from every other spot. What is more when the nearest spot is twice as far away, another spot which had been 5 times as far away is now 10 times distant. The speed of recession is proportional to distance.
Notice that our spots are discs set onto the surface and don’t change size. This is because galaxies (and stars and planets) are bound and do not increase in size with the cosmological expansion. Further Problems Dark Matter Each galaxy is attracted by gravity to every other galaxy and this should slow the rate of expansion. Will this gravitational attraction eventually stop the expansion and even ultimately lead to collapse? It depends upon the density of mass in the universe. In trying to assess this the mass of galaxies must be measured. Galactic rotations are found to be seriously anomalous but would fit our current understanding if we postulate a type of matter that does not interact with light – Dark Matter. There is estimated to be around 5 times as much dark matter as normal (proton, neutron electron) matter. Dark Energy Detailed measurement of the rate of expansion has revealed that rather than slowing down it is increasing. There is currently no generally accepted theory to explain this phenomenon but it has been named – Dark Energy.
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Section 7- Big Bang theory The temperature of stellar objects Stellar objects emit radiation From everyday experience we know that substances glow (incandesce) when heated to very high temperatures. The colour of light emitted depends on exactly how high the temperature is. When an object is heated it does not initially glow, but radiates large amounts of energy as infrared radiation. We can feel this if we place our hand near, but not touching, a hot object. As an object becomes hotter it starts to glow a dull red, followed by bright red, then orange, yellow and finally white (white hot). At extremely high temperatures it becomes a bright bluewhite colour. Early potters used the fact that materials glow different colours at different temperatures to determine the temperature inside their kilns. In 1792 the famous porcelain maker Josiah Wedgewood noted that all bodies become red at the same temperature. Potter’s guide Colour
Temperature °C
Dark red
550
Cherry red
750
Orange
900
Yellow
1000
White
1200
Black body Hot objects emit radiation with a continuous distribution of wavelengths (line spectra occur in special circumstances only with gases). An ideal surface that absorbs all wavelengths of electromagnetic (em) radiation is also the best emitter of electromagnetic radiation of any wavelength. Such an ideal surface is called a black body, ie a black-body is a perfect absorber and emitter of radiation. The continuous spectrum of radiation it emits is called black-body radiation. At room temperature this radiation is in the infra red range.
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The graph of radiated power versus wavelength has a characteristic shape. It is called a Planck distribution.
radiated power (arbritrary units)
30
25
20
1000°C 15
800°C 10
600°C 5
400°C visible light
0 0
500
1000
1500
2000
wavelength nm The black-body spectrum has three main features: ☼ The basic shape is more or less the same (apart from a scaling factor) at all temperatures, falling off gently on the long wavelength side of the peak, and much more sharply on the shorter wavelength side. ☼ As the temperature of the object is increased, the peak of the intensity spectrum shifts towards the shorter wavelengths. ☼ As the temperature of the object is increased, the power increases for all wavelengths. The properties of the spectrum are characterised by a single parameter, temperature, hence it is sometimes referred to as a thermal radiation spectrum. When the temperature is raised the peak moves towards the short wavelength, which gives the visible effect of changing from red to orange to yellow to white to blue-white, in that order.
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Stars Stars radiate electromagnetic waves like a black body. The peak wavelength of the emission spectrum tells us the surface temperature of the star. The star is classified by its temperature but it also gives an indication of its luminosity to use as a standard candle. Type
Surface temperature
Colour
Remarks
Example
O
≥ 33 000 °C
blue
Most of its em radiation is UV
several in Orion
B
10 000 – 33 000 °C
blue to blue white
Short lived stars
Rigel
A
7 500 – 10 000 °C
white
Common in our region of the galaxy
Sirius
F
6 000 – 7 500 °C
yellowish white
Common in our region of the galaxy
Capella
G
5 200 – 6 000 °C
yellow
8% of stars are this type
Sun, Polaris
K
3 700 – 5 200 °C
orange
12% of stars are this type
Arcturus, Alderbaran
M
≤ 3 700 °C
red
75% of stars are this type
Barnard’s Star
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Evidence for the Big Bang Big Bang theory From the expanding universe idea, if we consider running time backwards we come to the conclusion that the universe began from a single point called the initial singularity. The universe cools down as it expands During the very early stages of the universe, all matter and energy would have been concentrated in a much smaller volume. Its temperature would have been so high that electrons could not be bound with nuclei and the only state of matter would have been plasma. Plasma is opaque to em waves. As space expanded em waves are stretched, have longer wavelengths and hence less energy (see the Particles and Waves unit for an explanation of the energy of em waves). Eventually photon energy has reduced to allow electrons to bind with protons thus forming atoms (a gas) and thereafter em waves could travel freely through space. This early universe was equivalent to a black body with a temperature of about 3000 °C. The continued expansion of space stretched em waves such that now, 13 800 million years later, those waves are now found in the microwave part of the em spectrum with a peak wavelength at around 1 cm. Redshifting of light preserves the black-body spectrum while reducing the corresponding temperature. This “after glow of the Big Bang” is called the Cosmic Microwave Background radiation (CMB). ☼ CMB is almost identical from every direction. ☼ CMB corresponds to a temperature of radiation from an object at 2.725 K. (the Kelvin temperature scale starts at absolute zero - the temperature at which all atomic movement has ceased. 0 K is equal to -273.15 °C) ☼ Although CMB is very low in intensity, the universe is filled with it, so its total energy added up makes it the dominant radiation. History of the discovery of CMB radiation CMB was ‘discovered’ even before its existence had been predicted. It was discovered by accident. Theorists predicted its existence – not realising it had already been detected.
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COBE satellite The Cosmic Background Explorer (COBE) satellite was launched by NASA in 1989. It carried the Far-Infrared Absolute Spectrometer (FIRAS) experiment. More detailed observations of the CMB reveal further clues to its nature. For example, the radiation covers a range of frequencies with different intensities. By measuring the intensity of the radiation at various frequencies, a graph of intensity versus frequency can be produced. theory measurements
The spectrum indicates that the temperature of the universe is just a little above absolute zero, 2.725 ± 0.005 K, which is very close to that predicted. It is a more perfect black-body spectrum than any other known source. This provides absolutely incontrovertible evidence that the CMB was a black-body curve. Other evidence for the Big Bang The most abundant elements in the universe are hydrogen (70–75%) and helium (25–30%). The abundance of helium in the universe cannot be explained by it being synthesised inside stars. Most of it remains locked up in their cores. However, it can be accounted for by being formed during the first 17 minutes (or so) after the Big Bang using the physics which has been discovered in particle accelerators like the Large Hadron Collider at CERN. Bang? The Big Bang was not like a conventional explosion in which matter is thrown apart from a single point into a pre-existing empty space. Most cosmologists think space, time, matter and energy originated at the Big Bang and before that there was no space or time in the sense we think of them today.
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Rather than thinking of galaxies as flying away from each other through space, it is much better to think of them as being essentially at rest in an expanding space. A the idea steadily expanding universe leads to a number of paradoxes. Many of these can be overcome by postulating a brief (10 -32 s) extremely rapid expansion whereafter the Hubble’s constant rate governed. This concept is widely accepted and is called Inflation. Summary of evidence for the Big Bang theory ☼ The recession of the galaxies evenly in all directions. ☼ The cosmic microwave background radiation – a remnant glow of the hot early universe whose characteristics very closely fit those predicted by the expansion of space. ☼ The abundance of hydrogen, helium and the other elements throughout the universe agrees well with our understanding of the transmutation of light into matter as predicted by the theory. Endnote - The story of Cosmology Concept
Description
obsolete circa
Firmament
Fixed stars are points of light embedded in a dark sphere surrounding the Earth (planets orbiting inside it).
1600
Stars
Distant suns relatively evenly spaced throughout the infinite universe.
1930
Galaxies
Islands of 10 7 -10 14 stars relatively evenly spaced throughout the infinite eternal universe.
1970
Big Bang
Space expanding carrying galaxies away from each other from a primordial singularity.
In 1927 Georges Lemaître used Einstein’s equations to put forward a concept that the whole universe was expanding (based on the observation by Edwin Hubble of galaxies receding). Whilst the expansion was accepted by almost all astronomers this concept implied a finite universe with a definite start in time which many found difficult to accept (most notably Fred Hoyle the astronomer who coined the term Big Bang). They put forward rival ideas which permitted an eternal infinite universe of expanding space and were termed Steady State theories. The evidence that Steady State was wrong and overtaken by Big Bang theory became convincing by the 1970s. Interestingly it is now found possible that the universe may indeed be infinite and may even part of an eternal set of infinite universes - but these ideas currently find themselves in the realms of metaphysics.
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