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### Number 30

Gravity acts on all bodies that have mass. The gravitational force between two bodies is:

### •

always attractive;

### •

along the line joining the centers of mass of the two bodies;

### •

of the same magnitude on both bodies, but acting in opposite directions;

### •

proportional to the product of the two masses F m1 m2;

### •

inversely proportional to the square of their seperation F ∝ .

Newton’s Law of Gravitation states that the force between the attracting masses is :

1 r2

Worked Example. Two heavyweight boxers stand 1 metre apart. If they each have a mass of 80 kg, what is the gravitational force of attraction between them?

m1 = m2 = 80 kg.

F = newtons.

This is about 4.3 × 107 N, which is extremely small.

Worked Example. Calculate the gravitational force of attraction between one boxer (mass 80 kg ) and the Earth. Mass of Earth mE = 5.98 ××××× 1024 kg. Radius of Earth r

E = 6.36 ××××× 10 6 m.

This calculation is fairly straightforward but you should be careful with powers of ten, particularly when squaring distances in the denominator.

The calculation now looks like this: F = F = newtons F = 7.89 × 102 N Gm1m2 r2

### Gravitational field

A field is a region where we can detect a force. To find the gravitational field strength at a point, we place a mass m at the point and measure the (gravitational) force F.

The gravitational field strength is defined to be the force per kg. This ratio is also called the acceleration due to gravity g. Since g = F/m, the field strength may be quoted in Nkg−1, or ms−2

.

Field strength above the earth's surface

Consider a body of mass m at a point P a distance R from the centre of the Earth.

GmE m R2

Divide both sides by m gives the gravitation field strength g:

Gravitational field strength = = g = - an inverse square law.mF GmE R2

## Physics

The gravitational force exerted by the earth on a mass is the weight of the mass, which you will be familiar with from the formula W = mg, where g is the acceleration due to gravity.

This can be used to derive a relationship between g and G: Consider a mass m at the earth’s surface.

F = Gm1m2 r2

where G is the universal gravitational constant. (G = 6.67 × 10−11 Nm2kg−2). rE mE W = mg m F F r m1 m2

Exam Hint: - Ensure you measure the seperation, r, between the centres of mass of the two bodies, not between their surfaces.

6.67 × 1011× 80 × 80 12

6.67 × 1011× 80 × 5.98 ×1024 ( 6.36 × 106)2

We can write two expressions for the gravitational force acting on this mass due to the earth by using W = mg and Newton's Law of Gravitation. Equating these gives:

mg = Cancelling m gives:

g =

This approach can be used to calculate the acceleration due to gravity on any planet, by substituting the mass and radius of that planet for the mass and radius of the earth.

eg. Jupiter has mass 1.90 × 1027 kg and radius is 6.99 × 107 m. So the acceleration due to gravity on Jupiter is GmJ /rJ2 = 26ms-2

Gm mE rE2

GmE rE2

where mE= mass of the Earth rE = radius of the Earth

The force F is given by F =

m F P F mE R

(2)

Field strength below the earth's surface

For points below the earth's surface, the situation is slightly more complicated:

### Neutral point.

For any two bodies there is a point P in between them where the pull of one body to the right is balanced by the pull of the other body to the left. This is called a neutral point. A mass placed at P would be in equilibrium. Consider the planet Jupiter and its moon Ganymede:

F The graph below shows the variation in gravitational field strength (g)

with distance from the centre of the Earth. Note that the value of 9.81 that we often use is only correct at or near to the Earth’s surface - but even at the top of mount Everest, the reduction is less than 0.4% The field strength at point P is due only to the shaded sphere, of radius R, rather than the whole mass of the earth.

If we assume the density of the earth is approximately constant, the mass of the shaded sphere can be expressed as a fraction of the mass of the Earth by using: = So: = So: mS = mE R3 / r E 3

The gravitational field strength will therefore be: g =

=

So below the earth's surface, g is directly proportional to distance from the centre.

P R

mass of smaller sphere mass of Earth

volume of smaller sphere volume of Earth mS mER3/3rE3/3 GmE R3 / r E 3 R2 GmE R rE3 Worked Example

(a) Calculate the acceleration due to gravity at a height of 50 km above the surface of the earth.

Mass of Earth mE = 5.98 ××××× 1024 kg. Radius of Earth rE = 6.36 ××××× 106 m. G = 6.67 ××××× 10−−−−−11 Nm2kg−−−−−2

Use g = GmE/R2

Note that R = radius of earth + 50km = 6.41 × 106 m So g = 6.67× 1011 × 5.98 × 1024 /(6.41 × 106)2 g = 9.71 ms-2 rE 2rE 3rE 4rE

g ∝1 R2 9.81

### g/Ng

-1

(b) Calculate the height above the earth at which g is a quarter of its value on the surface of the earth.

We need to find the value of R when g = GmE/rE2 Surface value of g = GmE/rE2

At distance R from the center, we have g = GmE/R2 So we need to solve GmE/R2 = Gm E/4rE 2 Hence R2 = 4r E 2 So R = 2rE

So the height above the earth's surface is rE = 6.36 × 106 m Weightlessness is a term often used. However, as can be seen from the graphs and formula, there is always a force between the Earth and any mass and so the mass always has a weight. When you see pictures of orbiting astronauts it is better to describe them as apparently weightless. Field Lines

Field lines are used to show the presence of a gravitational field.

### •

The direction of the gravitational field lines shows the direction of the

force acting on a mass in the field.

### •

The spacing of the lines indicates the strength (or intensity) of the field. Lines close together = strong field; lines wide apart = weak field. The diagram below shows the gravitational field of a spherical mass such as the Earth. A field like this is called a radial field.

At increasing distance from the mass, the spacing increases showing that the field strength decreases - it follows the inverse square law detailed above.

P

Jupiter Ganymede

x

Since the gravitational force due to Jupiter is equal to that due to Ganymede at point P, we have

=

Substituting in and rearranging gives: y2 ( 1.90 × 1027 ) = x2(1.46 × 1023)

Square-rooting gives:

4.36 × 1013 y = 3.82 × 1011x

So 114y = x

Since x + y = Ganymede's orbital radius, we have 115y = 1.06 × 109,

giving y = 9.2 × 106 m GmJ x2 GmG y2 y

Ganymede's orbital radius:1.06 × 109 m

mG: 1.46 × 1023 kg

(3)

### Circular motion

Circular motion is covered fully in Factsheet 19, but to understand this section you just need to know that for a mass moving in a circle with constant speed:

### •

Angular speed ω (rad s-1), linear speed v (ms-1) and radius of circular path r (m) are related by the equation v = ωr

### •

The period of the motion (the time taken for one complete rotation) is given by 2π/ω

### •

There must be a force directed towards the centre of the circle (the centripetal force)

### •

The magnitude of the centripetal force is mω2r Consider the moon orbiting the Earth:

The force F which is necessary for circular motion of moons around planets, or planets around the sun, is provided by gravity.

Since the centripetal force is supplied by gravity, we have:

Hence: GmE =ω2

### R

3

Planetary system - Kepler's Laws

Applying the same principle as above to any of the planets in the solar system orbiting the sun, we obtain:

GmS = ω2 R3

where mS is the mass of the sun, and R is the orbital radius of the planet. But since the orbital period, T = 2π/ω, we have ω = 2π/T:

GmS =

R3

Rearranging: GmST2 = 4π2R3 This gives us: T2

### ∝

R3

This result is Kepler's Third Law of Planetary Motion.

NB: This derivation relies upon approximating the planets' orbits by circles, although they actually move in ellipses around the sun. However the result still holds true despite this.

### Gravitational Potential Energy

Consider a mass m in the Earth’s gravitational field.

ii) Again a = rE , but b = 3rE . (pe) = G mEm

(pe) = 6.67 × 1011× 6 × 1024× 1 × ∆ (pe) = 4.19 × 107 J 1 rE

3r1 . E

### [ ]

3 × 6.36 2 × 106 iii) Again a = rE , but now, b = . (pe) = GmEm

Note that = 0 and so,

(pe) = 6.67 × 1011× 6 × 1024× 1 × ∆ (pe) = 6.3 × 10−7 J or 63 MJ 1 rE

1 . 1

### [ ]

1 6.36 × 106

There is an important idea contained in this last part. The 1 kg mass has been taken to ‘infinity’ but the work done or gain in pe has increased by a finite amount ( 6.3 × 107 J). GmMmE R2 = mMω 2 R mE F mM R 2π T

### )

2 b ∆h a F mg m

To raise the mass from level a to b, the force F will do work F × ∆h on the mass, which increases the potential energy of the mass by this amount Since F = mg , the increase in potential energy is given by (p.e.) = mgh.

There are two important things to note here:

### •

We are only finding the increase in potential energy, we do not know the potential energy at a or b.

### •

We have assumed that g is constant between the levels a and b.

The assumption that g is constant is valid close to the Earth's surface, but for positions further away from the earth, another approach is needed.

Suppose we wish to move the mass between the two points from where R = a to R = b and find the work done for this displacement. Because the force F is varying we cannot use the equation work = F× d. Fortunately there is a simple way round the problem. First we find the work done in taking the mass from the point where R = a to infinity. For this displacement, the work done is

NB: This equation only works in a radial field.

To take the mass from the point where R = b out to infinity would require work of to be done.

So the change in potential energy between the two points is − = GmEm − mE F = G mE m R2 R = a m R = b R

### m

GmEm a GmEm b GmEm a GmEm b 1 a 1 b Worked Example

The Earth has a mass mE = 6.0 ××××× 1024 kg and radius r

E = 6.36 ××××× 10 6 m.

Find the gain in p.e. of a 1 kg mass when it is lifted from the Earth’s surface to a height of i) h = rE and

ii) h = 2rE.

iii) h = ∞∞∞∞∞ (infinity) Take G = 6.67 ××××× 10−−−−−11 Nm2kg−−−−−2

(pe) = G mEm (pe) = 6.67 × 1011× 6 × 1024× 1 × joules ∆ (pe) = 3.15 × 107 J 1 rE

2r1 . E

### [ ]

1 2 × 6.36 × 106 i) Using the above, we have a =rE and b = 2rE

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Lines of equipotential b a a q p Worked Example

A piece of space junk of mass 10 kg detaches itself from a spacecraft at p and falls freely to point q. Find the kinetic energy gained by the 10 kg mass in falling to Earth.

( rp = 30 ××××× 106m, r

q = 12 ××××× 10

6m m

E = 6 ××××× 10

24kg)

Since the mass falls freely from p to q, the gain in kinetic energy is equal to the loss in potential energy. The potential energies at points p and q are m × Vp and m × Vq.

where. Gain in ke = pe ( p ) – pe ( q ) = ( mVp ) – ( mVq ) = m (Vp Vq) = m = mGmE = 10 × 6.67 × 1011× 6 × 1024 = 2.00 × 108J VR = − GmE RGmErp GmE rq

r1 . q 1 rp

30 ×1 106

. 1 12 × 106

### ( )

Defining Gravitational Potential Energy

To define the actual potential energy, rather than working out differences in it, it is first necessary to decide at which point a body will have zero potential energy. At first sight it may seem strange but it has been agreed that this point shall be at ‘infinity’.

Since moving a point from the earth's surface to infinity increases its potential energy (as seen in the previous example), if the potential energy is zero at infinity, it must be negative elsewhere.

Gravitational potential energy of a mass at a point in a gravitational field is the work done in bringing the mass from infinity to that point.

Gravitational Potential (V)

Gravitational potential (V) is the gravitational potential energy per unit mass.

(Jkg1) GmE

R V =

The diagram below shows how the potential varies with the distance from the earth’s centre.

### Typical Exam Question.

(i) Use Newton’s law of gravitational to find the units for G (ii) Hence show that the expression

does have the unit joule.

(i) Newton’s law: re-arrange

Substituting the known units, we obtain the units for G : [G] = or Nm2kg2

(ii) pe = . Substitute the units: [pe] =kg × (Nm

2kg2) × kg m

= (−)Nm.

Since work (or energy) is force × displacement it follows that Nm is the same as joules.

F = Gm1m2 r2 G = Fr2 m1m2 Nm2 kg2 GmE Rm ×

R = distance from centre of earth (R >rE)

To find the potential energy of a body at a given point, multiply the potential at the point by the mass of the body.

V/107 Jkg−1 0 −2 −4 −6 Earth 1 R V ∝ 5rE 10rE 15rE R GmE R −−−−− m ××××× Lines of equipotential

These are lines which surround a body, all points on the line being at the same potential. They are shown as dotted lines in the diagram below.

Note that the lines of equipotential are perpendicular to the field lines. If you move along the curved path from a to b then the displacement is always perpendicular to the direction of the force. This means no work is done and there is no gain (or loss) in potential energy; hence the curve through ab is a line of equipotential.

### Escape Velocity

The escape velocity is the velocity a mass must have to escape from the Earth ( or any celestial body) and not return. The mass must not be driven by motors of any kind, but have a velocity imparted to it that will carry it completely away from the Earth ie to infinity. As the mass gets further from the Earth, its potential energy increases. At infinity, the potential energy has increased to zero. When the mass has its escape velocity u, it must have sufficient kinetic energy to match the necessary gain in potential energy in taking it to infinity. Let u = the escape velocity for the Earth. We use conservation of energy for a mass m.

Loss in kinetic energy = Gain in potential energy. ½ mu2 = pe() pe(r E) ½ mu2 = 0

2GmEm

### ]

rE2GmEm r = 2 × 6.67 × 10−11 × 6 × 1024 6.36 × 106 = 11 × 10 3 ms−1 u = Lines of Field

(5)

a) Write down an expression for the gravitational potential at a distance R from the centre of a planet of mass M. Hence write down an expression for the potential energy of a mass m at a height h above the Earth’s surface ; radius of Earth = RE , mass

Earth = ME . [3]

Potential

Potential energy = mgh. ! 1/3

b) An amount of energy E E E E is transferred to a satellite of mass mE when launched by a certain rocket from the surface of the Earth. If the vertical height reached by the satellite is h, obtain an equation relating h to E E E E E and the other terms. [2] E = mgh = m h. ! 0/2

c) Show that if = RE, the energy E E E E is given by EEEEE = [2]E

Putting RE = h, E = m ! 0/2

d) The graph below shows how the force acting on the satellite varies with distance from Earth’s centre.

### Exam Workshop

V = GM R

Minus sign omitted. Has not understood how to find potential energy from potential . Formula quoted only valid when h « RE . In this case h = RE .

GME RE2

A fatal error has been carried forward and gains no credit. Substituting for g = is meaningless as this expression is only true at the Earth's surface. This candidate has relied too much on quoting formula without understanding the physics.

GME RE2 GMEm 2RE GME RE

Write the equation that shows how this force can be calculated and given that the force at the Earth’s surface is 10 kN, calculate the force on the satellite at a height h = RE. [3]

Force on satellite /kN RE 2RE 3RE 4RE 0 2 4 6 8 10

Distance from Earth centre

The force follows Newton's Law of Gravity.

2

R Mm G F=

When h = R, the distance is doubled so

### ( )

2 2 R Mm G 4 1 2R Mm G F= =

The force is reduced to a quarter, so force at height R is: 2.5kN 4 10kN = ! ! ! ! ! ! !! ! ! ! ! ! ! !

This answer is correct and gains full credit. It is testing the inverse square law.

e) Indicate on the graph the quantity that represents the energy E.E.E.E.E. [2] Force on satellite /kN RE 2RE 3RE 4RE 0 2 4 6 8 10

Distance from Earth centre ! !! !! ! ! ! ! !

This answer is correct and gains full credit. It is testing how to find the work done by a force which varies with distance.

This is a general result and is used elsewhere in physics.

a) R GM V potential =− E

potential energy of mass m = m × V, at height h, R = R E + h and so p.e. =

### )

h R GM m E E + × −

b) Energy E is transferred so by conservation,

E + − =−

+

### )

h R GM m R GM m E E E E c) When h = R E we get E + ⇒     − =     − E E E E 2R GM m R GM m E = E E 2R GM m

d) As above, force follows inverse square law:-

2

R mM G F=

Doubling distance gives

4 1 2 1 2 =       of force. Force when h = RE 2.5kN 4 10 =

e) Shade area under graph from RE to 2RE

! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! ! ! !!!!! ! ! ! ! ! ! !! !! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !! !! !!!!!

(6)

### Typical Exam Question

1) i) Explain the term ‘gravitational potential’ as applied to a point in the Earth’s gravitational field.

ii) The gravitational potential at the Earth’s surface is

−−−−−63 MJ kg-1 ; taking the Earth to be a sphere of radius R

what are the potentials at points A and B at heights R and 3R respectively above the Earth’s surface?

iii) A 10 kg mass is moved from A to B. Does its potential energy increase or decrease and by how much ?

iv) A 1 kg mass at the Earth’s surface is taken to infinity. What is the minimum amount of energy to achieve this ?

i) The gravitational potential at a point is the work done per kilogram by an external force in moving a mass from the point to infinity (or zero potential) Or, alternatively, the work done per kilogram by the field as the mass ‘falls’ from infinity

to the point. [2]

ii) At the Earth’s surface radius R, the potential is given by

and the potentials at A and B are given by

VA = 31.5MJ kg-1 and V

B =15.75 MJ kg

-1 [4]

iii) The potential energy is m × V joules. So, for the 10 kgmass at points A and B, we have: p.e. (A) = 10 × (31.5) MJ = 315 MJ p.e.(B) = 10 × (15.75 ) MJ = 157.5 MJ The potential energy at B is less negative than at A, hence in moving from A to B, the potential energy increase [3]

iv) The potential at infinity is zero, hence the potential energy of any mass at infinity is zero. The potential energy of 1 kg mass at Earth’s surface is –63 MJ. To move it to infinity the potential energy must increase to zero. Hence, the minimum energy

required is 63 MJ. [3] V = GM= −63MJ kg-1 R VA = GMand VB = − and so 2R GNM 4R VA = MJ kg-1 and V B = MJ kg -1 62 2 63 4 ! ! ! ! ! ! ! ! ! ! ! !

(Where necessary, take G = 6.67 × 10-11 Nm2kg-2 )

1 a) Explain the term ‘gravitational field’ [2] b) Define the term ‘gravitational field strength’ [2] c) A communications satellite orbits the Earth in a circular path of

radius 4.2 ×107 m. Calculate the gravitational field strength at a

point in this orbit (Mass of earth =6×1024kg) [3 ]

d) The figure below shows a spherical planet with some equipotentials.

### Questions

a d b c -1500 Jkg-1 -1000 Jkg-1 -700 Jkg-1

### ·

0 -10 -20 2 4 6 8 R/I V/kJ kg-1 X Y

iii) Find the work done in moving the 10 kg mass from point X to point Y (6×104m from planet’s centre). [2]

iv) Write down the gravitational potential at point X and hence deduce the gravitational field strength (intensity) at point X. What force acts on the 10kg mass? [2 + 1] 2 a) Define the term ‘gravitational potential’ at a point in a radial

gravitational field.

b) Use this definition to calculate the ‘escape velocity’ for the Earth with mass ME = 6×1024 kg and radius R

E = 6.36×10 6 m.

c) A 1kg mass is to be catapulted from the Earth’s surface so that it will not return. How much energy must be given to the mass? What is this energy in kilowatt hours and how much would it cost (at the domestic rate of 5.5pence / kWh) to expel this mass? d) The planet Mercury is smaller than Earth with a mass

MM = 3.24 ×1023 kg and radius R

M = 2.34 × 10 6 m.

Calculate the escape velocity for Mercury and, giving reasons, comment on the likelihood of Mercury retaining any atmosphere. Assume G = 6.67 × 10-11 Nm2kg-2

e) i) Write down an expression for the gravitational potential at a distance R from the centre of a planet of mass M. [2] ii) Using the graph below find the potential energy of a 10 kg mass

at 4×104m from the planet’s centre (point X). [2]

Calculate the gain in potential energy when a 10 kg mass is moved i) from a to b

ii) from b to c

3. a) Write down Newton’s law of gravity for two attracting masses. b) Assuming the Earth to be a uniform sphere of mass M and radius

R show that the acceleration of free fall at the Earth’s surface may be deduced from Newton’s law and is given by:

c) Given that the mean density of the Earth is ρ, show that

d) Show that the units on both sides of this equation are consistent. e) A spaceship hovering above the surface of Mars releases a mass at a height of 190m . This mass hits the surface of the planet at 38ms-1.

Calculate a value for the acceleration of free fall at the planet’s surface. f) Given that Mars has a radius of 3.4 × 106 m, estimate a value for the

density of the planet. Assume G = 6.67 × 10-11 Nm2kg-2 g = GM R2 3g G = 4πρR

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1. a) Space around a mass where gravitational effect (force on second mass ) can be detected.

b) Gravitational field strength /intensity at a point defined as force per unit mass at that point =

c) = 0.227 Nkg-1 (ms-2) d) pe = m ×V gain pe(ab) = (-10×1000) − (-10×1500) = 5000J gain pe(bc) = (-10×700) − (-10×1000) = 3000J gain pe (dc) = zero. e) i) potential

ii) from graph, at X, potential = -11.5kJkg-1

⇒ pe(x) = -11.5×10 kJ = -115 kJ

(allow for sensible error in graph e.g. 11<V<12 ) iii) pe (Y) = -7.5×10 kJ. Work done = increase in pe

= (-75kJ) –(-115kJ) =40kJ. iv) For sphere,

at X, = 2.875 ms-2 2.9 ms-2

force = 10 × 2.9 = 29 N

2. a) work done per kg in bringing mass from infinity to the point. b) Escape = = 11.2 × 103 ms-1

c) = 6.3 × 107 joule 1kWh = 3.6 × 106 J k.e. = 17.5 kWh

Cost = 96.25 pence ! d) uescape= 4.3 × 103 ms-1

smaller speed, easier escape, atmosphere less likely. (See also kinetic theory; molecular k.e. ∝ Temperature)

3. a) & b) see text

c) put M = ρ × volume = ρ×4πR3 /3

d) sub in units, change N to kg × ms-1

e) g = 3.8ms-2

f) ρ = 4.0×103 kgm-3

Acknowledgements:

This Factsheet was researched and written by Keith Cooper

Curriculum Press, Unit 305B, The Big Peg, 120 Vyse Street, Birmingham, B18 6NF Physics Factsheets may be copied free of charge by teaching staff or students,

provided that their school is a registered subscriber.

No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

F1 and F2 are mutual gravitational attraction. F1 = F2 ii) centripetal force = F2

iii) Kepler’s third law : T2 R3

iv) GM = R3ω2

i.e. Kepler’s third law : T2 R3a GMm R2 = m × Rω 2 ω = = GMT2π 2 = 4π2R3 T i) ! ! ! ! ! ! !

2) A moon of mass m is orbiting its planet of mass M in a circular orbit of radius R.

i) Draw a diagram showing the forces acting on both bodies. How do these forces arise and are they equal in magnitude

or not ? [2]

ii) Which of the two forces you have drawn provides the centripetal force necessary for circular motion ? The moon orbits the planet in a time T. [1] iii) State Kepler’s third law of motion as applied to the moon

[2] iv) Show how Kepler’s third law may be deduced from

Newton’s law of universal gravitation. [3]

M

F1 F

2

m

### Typical Exam Question

F m GM R V = − 6.67× 10-11× 6× 1024 F = GmM R2F m = GM R = (4.2× 107)2 V= G mM R2V R = GM R2 = g -11.5× 103 = 4× 104 g R 2GM k.e.=mu 2 2

References

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