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By Mrs Lim 2
What do you know about gravitation? 3
• It caused an apple (lucky not durian!) to fall on Newton’s head about 400 years ago. 4
ü 4 Jan 1643 – 31 March 1727 ü In Manor House at Woolsthorpe, Lincolnshire ü Educated at Grantham Grammar School, Lincolnshire 5
cool isn’t it? A one pound note with Newton’s portrait! (Imagine one with your portrait) 6
Gravitational Force is …. ü attractive in nature. ü exerted by stars and planets. (As well as the apple) ü Gravity is stronger on Earth than on Moon. ü Gravitational effects diminishes with distance…. 7
1. 2 Newton’s Law of Gravitation The Newton’s law of gravitation states that every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. 8
F F m 1 m 2 r 1. - equal and opposite attractive force on each other 2. - directed along the line joining the two particles. 3. - for a sphere of uniform density, the force that it exerts on other objects can be obtained by treating the sphere as a point mass at the centre of the sphere. 9
F F m 1 m 2 r Qn: What should we do if not spherical or not point mass? Ans: The formula is still applicable if the objects are placed sufficiently far apart such that their sizes become negligible compared to the distance separating them. 10
Quick Check 1 Earth and the Moon are gravitationally attracted to each other. Does the more massive Earth attract the moon with (a) a greater force (b) the same force (c) or less force than the moon attracts Earth? Ans: (b) 11
Quick Check 2 The weight of an object on peak of Himalayas ( about 8848 m above sea level) is (a) the same as (b) slightly smaller than (c) much smaller than the weight of that object at sea level. Ans: (b) 12
Quick Check 3 n A) B) C) D) On the ground the gravitational force on an object is W. What is the gravitational force at a height R, where R is the radius of the earth? 0. 25 W 0. 5 W W 2 W Ans: [ A ] 13
Explanation: 14
Worked Example 1 (in tutorial worksheet) Assuming that they can be treated as point masses, find the gravitational force between (a) two persons of masses 60 kg and 1. 0 m apart, (b) a person of mass 60 kg at the surface of the Earth, and the Earth of mass 6. 0 1024 kg and radius 6400 km. (a)Force = = = 2. 4 10– 7 N 15
Example 1 Assuming that they can be treated as point masses, find the gravitational force between (a) two persons of masses 60 kg and 1. 0 m apart, (b) a person of mass 60 kg at the surface of the Earth, and the Earth of mass 6. 0 1024 kg and radius 6400 km. (b) Force = = = 590 N 16
TWO NEW CONCEPTS • Gravitational Field Strength 17
Gravitational field is a region in which a mass experiences a gravitational force. 18
2. 2 Gravitational Field Strength The gravitational field strength at a point is defined as the force per unit mass acting at that point. Field Strength at a point g = FG / m Unit: N kg-1 19
Field Strength of the gravitational field of a point mass or spherical mass M r P Gravitational force experienced by a small mass ms placed at point P is 20
Field Strength of the gravitational field of a point mass or spherical mass Field strength at point P is g = = = 21
Field Strength of the gravitational field of a point mass or spherical mass • Minus sign indicates attractive nature of the force (rather than repulsive). • NEED NOT be included during calculations • Usually included in graphical representation of F and g. 22
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Quick Check 4 (a) (b) (c) The gravitational field generated by (or due to) the more massive Earth will be a greater the same or less than the gravitational field generated by the moon? Ans: (a) 24
Quick Check 5 The gravitational field experienced by an object on the peak of Himalayas ( about 8848 m above sea level) due to the Earth is (a) (b) (c) the same as slightly smaller than much smaller than the gravitational field experienced by the object at sea level. Ans: (b) 25
Quick Check 6 Karen’s mass is 50 kg and Kenny’s mass is 75 kg. They are both sitting in the TPJC Auditorium. (Considering only their interaction with Earth) They will experience the same A. Gravitational force B. Gravitational field strength Ans: (B) 26
Worked Example 2 (in tutorial) The mass and radius of the Earth is 6. 0 1024 kg and 6. 4 106 m respectively. (a) Find the gravitational field strength at (i) the surface of the Earth, (ii) 3000 km above the Earth’s surface. (b) What is the assumption made in the calculations? (a) (b) (i) At Earth’s surface, distance from centre of Earth, r = radius = 6. 4 106 m 9. 8 N kg– 1 27
(a) (ii) At a height 3000 km above Earth’s surface, distance from centre of Earth, r = 6. 4 106 + 3000 103 = 9. 4 106 m 4. 5 N kg– 1 (b) In our calculations, we have assumed Earth as a uniform sphere. 28
Quick Check 7 If a third object sits between two massive objects like (say when the moon orbits round the earth, it will experience the Sun’s gravitational field as well. ) What will be the total gravitational field? How should they be added? (A) as a scalar (B) as a Vector ANS : (B) 29
– look at worked example 3 in tutorial 30
2. 5 Graphical representation of Earth’s field strength Earth M field strength g RE distance r Why? (see appendix 6) The formula is only valid for r > RE For r < RE the relation is linear? 31
2. 6 Acceleration of free fall Field Strength g = F/m Acceleration of a free falling object a So a = = F/m g Gravitational field strength of the Earth is 9. 81 N kg-1 (or 9. 81 m s-2) near its surface and is approximately constant due to the large size of the Earth. 32
SUMMARY Field Strength at a point g = FG /m (Definition) (M being a point mass or spherical mass) 33
A closer look at gravitational potential energy. . . 34
What does gravitational potential energy depends on? • It depends on the mass of the object • It depends on the position 35
3. 2 Formal Definition of Gravitational Potential Energy The gravitational potential energy of a mass at a point is defined as the work done by an external agent in bringing the mass from infinity to that point. 36
displacement M P gravitational force external force m at infinity • The work is done by an external force acting in a direction opposite to the gravitational force. • The direction of external force is opposite to the displacement of the mass. Hence the work done by the external force is negative. • There is no change in velocity or kinetic energy. 37
It can be shown that for two point or spherical masses attracting each other with gravitational force (look at lecture notes) 38
or Potential Energy = mgh ? Oops so which equation is correct? 39
Lets Explore: A 3 -kg mass is projected to a height h above the Earth’s surface. Use U = mgh and to find the change in gravitational potential energy if (i) h = 1000 m (ii) h = 1000 km Radius of the Earth is 6400 km; Mass of Earth = 6 x 1024 kg. 40
(i) Using U = mgh, U = 3 x 9. 8 x 1000 = 2. 94 x 104 J Using So there is not much difference between the two formulae for h = 1000 m. 41
(ii) Using U = mgh U = 3 x 9. 8 x 1000 = 2. 94 x 107 J Using This time there is a significant difference. U = mgh is inaccurate in this case. Why ? 42
This is because the acceleration due to free fall (in other words “g” on the object) is not constant, hence the equation mgh will not be accurate. (see “A” level Physics is DIFFERENT from “O” level) 43
3. 3 PROJECTION OF A BODY FROM SURFACE OF A PLANET low projection velocity high projection velocity The maximum height reached is obtained by equating PEgained = ½mv 2 - 0 44
Graphical Representation Potential Energy EK RE dist. from centre of Earth r 0 ET = Ep + Ek EK 45
Condition for Escape to Infinity v to infinity rocket Earth k. e. 0 k. e. = ½mv 2 p. e. = 0 p. e. = -GMm/r Note that at infinity, total energy 0 By conservation of energy, the condition for escape is. . . 46
K. E. + P. E. i. e. ½mv 2 + (-GMm/r) v 0 0 So the escape speed is 47
SUMMARY • Increase in GPE is work done against gravity. • Formal definition of GPE (must remember). For two point or spherical masses attracting each other with gravitational force, • This formula gives the absolute potential energy with infinity set as the reference point (i. e. GPE = 0) 48
4 GRAVITATIONAL POTENTIAL The gravitational potential at a point is defined as the work done per unit mass by an external agent in bringing a mass from infinity to that point. Based on this definition, we get and hence for point mass or spherical mass, 52
For example, we use a 1 kg mass…. W. D. = - 6 J OR U = - 6 J = -8 J kg-1 Infinity W. D. = - 2 J U = - 2 J OR =0 = - 2 J kg-1 How will the workdone change if the mass of the object is 2 kg? 53
Worked Example 5 X and Y are two points at respective distances d and 3 d from the centre of the Earth, where d is greater than the radius of the Earth. The gravitational potential at X is – 900 k. J kg– 1. What is the work done on a 2 kg mass when it is taken from X to Y? Solution Since potential -1/r, so -300 k. J kg– 1 WD = m ( Y - X) = 2 [ (– 300) – (– 900) ] = 1200 k. J 54
5 RELATION BETWEEN FIELD STRENGTH AND POTENTIAL Verify, for point mass, that gravitational potential distance r gradient gives field strength 55
Similarly These two equations are ALWAYS correct. 56
Worked Example 6 The following table gives the values for the gravitational potential due to the Earth at the various heights above the Earth’s surface. Height / km 580 600 620 Gravitational potential / MJ kg– 1 -57. 51 -57. 34 -57. 17 Deduce a value for the Earth’s gravitational field strength at a height of 600 km. 580 600 620 r 57
Solution Field strength = = – 8. 5 N kg– 1 Magnitude of field strength is 8. 5 N kg– 1 The minus sign indicates that the direction of field strength is towards the Earth’s centre. 58
In the auditorium, an invisible friend is also attending lecture and he is quite worried after learning about the attractive gravitational force…. . If Gravitation force (due to the Earth) acts on the moon; then the Moon should collapse onto the Earth…. 59
THE SKY IS FALLING!!!!! 60
Do You agree? Since there is attractive gravitation force exerted by the Earth on the Moon and since the moon is less massive, it should move and collide with the earth …Since the Sun has the same effect on the Earth, then Earth should move and hit the Sun…. Then what is left? ……. Is the sky falling ? Is it the end of the world ? 61
6 ORBITS • There are many naturally occurring orbital motions. • Gravitational forces between heavenly bodies are keeping them in their curved orbits. • Most orbits are elliptical, but we will only consider circular ones. • In circular orbital motion, the gravitational force acts as the centripetal force. 62
Johannes Kepler 63
r Gravitational force = F centripetal force 64
Worked Example 7 (a) Assume that the planets move in circular orbits about the Sun, show that the squares of their periods of revolution are proportional to the cubes of the radii of their orbits. (b) The Earth is 1. 50 1011 m from the centre of the Sun and takes exactly one year to complete one orbit. The planet Mars takes 1. 88 years to complete an orbit of the Sun. Calculate the radius of Mars’ orbit. 65
Solution (a) gravitational force = T 2 centripetal force r 3 66
(b) Since T 2 r 3, then r. M = 2. 28 1011 m 67
6. 2 What about energies associated with a satellite? Imagine a satellite of mass m moves in a circular orbit about the Earth of mass M. The radius of the orbit is r. We’ll derive an expression for (a) the kinetic energy T of the satellite, (b) the gravitational potential energy V of the satellite, (c) the total energy E of the satellite. 68
(a) Gravitational force = centripetal force kinetic energy, T = 69
(b) Gravitational potential energy, V = (c) Total energy, E = T + V Note: A negative total energy indicates that the satellite does not have enough energy to escape from Earth’s gravitational field to infinity. 70
6. 3 The GEOSTATIONARY (Geosynchronous) Orbit A Geostationary orbit is one in which a satellite orbiting in it will always appear as stationary when viewed from any point from the Earth. It has the following characteristics: • Period = one day ( 24 hours). • The whole orbit is in the equatorial plane. • The direction of revolution is the same as the direction of Earth's rotation. • Since T 2 r 3, it also has a particular radius. 71
Application… n If you own a communication empire and your company will be launching a new geostationary communication satellite. What would be the radius of the orbit? 72
Find the radius of the geostationary orbit. gravitational force = centripetal force 73
For geostationary orbit, we substitute T = 24 hours = 86400 s, M = 6. 0 x 1024 kg, and obtain The radius of a geostationary orbit is 4. 23 107 m. 74
Summary 75
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Orbital Motion (a) gravitational force = centripetal force 77
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