Innova Junior College
JC1 Physics
Topic 8 Gravitational Field
Gravitational Force and Field
1. How far from the Earth must a body be along a line toward the sun so that the sun’s
gravitational pull balances the Earth’s? The sun is 1.49 x 108 km away from the Earth and its
mass is 3.24 x 105 times the mass of the Earth.
[2.61x 108 m]
FS
Sun
FE
Earth
Let r be the distance between the body and the centre of the Earth such that FS = FE.
FS  FE
GMS m GME m

r2
r '2
G  3.24  105 ME  m G  ME  m

2
r2
1.49  1011  r 
r = 2.61 × 108 m
2. J2000/I/7
Outside a uniform sphere of mass M, the gravitational field strength is the same as that of a
point mass M at the centre of the sphere. The Earth may be taken to be a uniform sphere of
radius r. The gravitational field strength at its surface is g. Show that the gravitational field
2
strength at a height h above the ground is gr .
r  h
2
On earth’s surface [a distance of r away from the centre of Earth]
GM
g 2
r
GM = gr2
At h above ground [a distance of (r + h) away from the centre of Earth]
GM
gr 2
g 2 
2
r'
r  h
3. N02/II/1
Towards the end of the eighteen century, Cavendish successfully measured the force of
attraction between two spheres. Hence, he became one of the first scientists to determine a
value for the gravitational constant G.
(a) Write down an equation representing Newton’s law of gravitation, explaining any
symbols used.
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(b) The acceleration of free fall g at the Earth’s surface is given by the expression
GM
g 2
r
where M is the mass of the Earth and r is its radius.
Calculate a value for its mass of the Earth, given that r = 6.40 × 106 m. [6.02 × 1024 kg]
(c) Some textbooks describe Cavendish’s experiment as a means of ‘weighing the Earth’.
Suggest why this statement is incorrect.
GMm
r2
where FG : gravitation force of attraction between two masses M and m
G : gravitational constant
M and m : the masses of the two particles (spheres)
r : distance between the centres of the two masses
(a) FG 
GM
r2
At the surface of the Earth (where r = 6.40 × 106 m), g = 9.81 m s-2
6.67  1011M
9.81 
2
 6.40  106 
(b) g 
M = 6.02 × 1024 kg
(c) By describe Cavendish’s experiment as weighing Earth, the textbooks are implying the
experiment is measuring the force exerted on Earth by another object. However, his
experiment is a mean of measuring the mass of the Earth.
Gravitational Potential and Potential Energy
4
J80/II/5 (modified)
X and Y are two points at respective distances R and 4R from the centre of the Earth, where
R is greater than the radius of the Earth. The gravitational potential at X is – 800 kJ kg-1.
What is the work done on a mass of 2.5 kg when it is taken from X to Y?
[+ 1.50 × 106 J]
GM
  8.00  105
R
GM  800  103
Y  

  2.00  105
4R
4
W  U
 m
X  
 2.5   2.00    8.00   105
= + 1.50 × 106 J
Note: the work done on the mass is purely for the change in gravitational potential energy
due to the change in position in the gravitational field.
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5
J88/I/18
A stationary object is released from a point P a distance 3R from the centre of the Moon
which has radius R and mass M.
Moon
P
R
3R
Determine the speed of the object, in terms of G, M and R, on hitting the Moon.
When the object is released from rest, it will move from high potential to low potential (i.e. P
to surface). In doing so, the object will lose gravitational potential energy and gain kinetic
energy.
Using conservation of energy,
GPEP + KEP = GPEM + KEM
 GMm 
 GMm 
2

  0  
  ½ mv
3
R
R




4GM
v
3R
Using the idea of dynamics:
The stationary object is subjected to a gravitational force that will attract it towards the Moon. Thus it
will accelerate towards the Moon, gaining velocity. However the acceleration is not a constant value
since the gravitational force acting on the object is not constant. Gravitational force varies inversely
proportional to the square of the distance, r between the centre of the objects. As the object moves
towards the Moon, the distance r decreases with gravitational force increasing. Hence with a nonconstant resultant force, equation of motion, in particular equation v2 = u2 + 2as is not appropriate for
this calculation.
6
J85/II/9 (part)
The curve in the graph shows the way in which the
gravitational potential energy of a body of mass m in the
field of the Earth depends on r, the distance from the
center of the Earth, for values of r greater than the Earth’s
radius RE.
What does the gradient of the tangent to the curve at
r = RE represent?
The body referred to above is a rocket which is projected
vertically upwards from the Earth. At a certain distance R
from the centre of the Earth, the total energy of the rocket
(i.e. its gravitational potential energy plus its kinetic energy)
may be represented by a point on the line PQ. Five points
A, B, C, D, E have been marked on this line.
Which point (or points) could represent the total energy of
the rocket
(a) if it was momentarily at rest at the top of its trajectory,
(b) if it was falling towards the Earth,
(c) if it was moving away from the Earth, with sufficient energy to reach an infinite distance?
In each case, explain briefly how you arrive at your answer.
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The gradient of the tangent of a gravitational potential energy against distance curve at r =
RE is the gravitational force, i.e. weight of the body at the surface of the Earth.
dU
FG  
dr
(a) if the rocket was momentarily at rest at the top of its trajectory, its velocity and hence its
kinetic energy is zero. Hence the total energy only consists of U. Therefore point B could
represent the total energy of the rocket.
(b) Since the rocket is falling back to Earth, it is still bound to Earth, hence total energy must
still be a negative value. In addition, when the rocket was falling, its velocity is of a
particular value and hence the rocket has a particular (positive) kinetic energy. With its
total energy being more than its gravitational potential energy and being a negative
value, point C could represent the total energy of the rocket.
(c) With sufficient energy to reach infinite distance, the rocket is no longer bound to Earth.
Hence its total energy should be more or equal to zero. Point D and E could represent
the total energy of the rocket.
7
J01/III/2 (part)
b The moon Charon (discovered in 1978) orbits the planet Pluto. The graph shows the
variation of the gravitational potential  with distance d above the surface of Pluto along
a line joining the centres of Pluto and Charon.
The gravitational potential is taken as being zero at infinity.
(i) Suggest why all values of gravitational potential are negative.
(ii) Use the graph to determine, giving an explanation of your working,
1. the distance from the surface of Pluto at which the acceleration of free fall is
zero,
2. the acceleration of free fall on the surface of Charon.
c
A lump of rock of mass 2.5 kg is ejected from the surface of Charon such that it travels
towards Pluto.
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(i) Using data from the graph, determine the minimum speed with which the rock hit the
surface of Pluto.
(ii) Suggest why, if the rock travels from Pluto to Charon, the minimum speed on
reaching Charon is different from that calculated in (i).
b
(i) Since the direction of external force and that of the displacement from infinity to the
point of concern is opposite in direction, work done by the external force, and hence
gravitational potential is negative.
(ii) When the resultant force only consists of gravitational force, Fn = ma, i.e. mg = ma.
a=g
Thus the acceleration of free fall is equivalent to the gravitational field strength which
is reflected by the gradient of a gravitational potential – distance graph.
1. when acceleration of free fall is zero, the gradient to the gravitational potential
verus distance is zero. This occurs at a distance of 13.6 × 106 m.
2. The acceleration of free fall on the surface of Charon:
aC≡ potential gradient at the surface of Charon
1.35  10 6

2.6  10 6
= 0.052 m s-2
c
(i) In order for the rock to reach Pluto, the rock must be given kinetic energy that is
equivalent to the gain in gravitational potential energy to reach the point with zero
gravitational potential gradient. From there, it will move to the surface of Pluto,
converting its total energy (only comprises of gravitational potential energy) to kinetic
energy.
GPEC + KEC = GPEmax + KEzero = GPEP + KEP
 2.5  29.565  106   0   2.5  30.0  106   ½ 2.5  v 2
v = 933 m s-1
(ii) For the flight of the rock from Pluto to Charon, the kinetic energy gained is the loss in
gravitational potential energy from the point with zero gravitational potential gradient
to the surface of Charon. This loss in gravitational potential energy is much lower
than when the rock move from the point with zero gravitational potential gradient to
the surface of Pluto. Thus the minimum speed on reaching is lower than that
calculated in c(i).
Satellite Motion and Circular Motion
8
J93/II/1 modified
A planet P of mass m orbits the Sun S of mass M in a circular orbit of radius r and period T
as shown on the diagram below.
P
S
r
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(a) On the diagram, draw an arrow representing the linear velocity of P and label this v.
Draw a second arrow representing the direction of the resultant force acting on P. Label
this F.
(b) (i) Write down an expression, in terms of m, r and T, for the magnitude of resultant force
F.
(iii) Write down an expression, in terms of m, M, r and G, for the magnitude of the
gravitational force exerted by the Sun on the planet.
(iii) Hence, show that T 2  r 3 .
(a)
v
P
F
S
r
4 2 mr
 2 

(b) (i) Fc  mr 

T2
T 
Mm
(ii) FG  G 2
r
(iii) FC  FG
2
4 2 mr GMm

T2
r2
4 2 3
T2 
r
GM
T2  r3
9
N79/I/4 (part)
Explain what is meant by (i) gravitational field strength, (ii) gravitational potential. How do
these quantities vary with distance x from a point mass?
It can be shown that, for planets performing circular orbits about a sun, the period of
revolution  is related to the radius r of the orbit by the equation  = Arn, where A and n are
constants. Values of  and the mean radius r for some of the planets of our solar system are
given in the table below.
Planet
r / 106 km
 / year
Mercury
Venus
Mars
Jupiter
Saturn
0.241
0.615
1.88
11.9
29.5
58
108
228
778
1430
a. By drawing a suitable graph, deduce the values of A and n.
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b. Find the mean radius of the Earth’s orbit.
Gravitational field strength, g, at a point in free space is defined as the force per unit mass
acting on any object placed at the point.
Consider g = GM/x2, g decreases as x increases.
Gravitational potential, , at a point due to the gravitational field set up by a mass M is
defined as the work done by an external force in bringing a unit mass from infinity to that
point.
Consider  = - GM/x,  increases as x increases.
(a) Given:   Ar n
Consider lg   lg A  n lg r
Planet
Mercury
Venus
Mars
Jupiter
Saturn
/year
0.241
0.615
1.88
11.9
29.5
r/106 km
58
108
228
778
1430
lg (/year)
-0.618
-0.211
0.274
1.076
1.470
lg (r/km)
7.76
8.033
8.358
8.891
9.1553
Plot lg  against lg r
Gradient of graph, n
1.42  ( 0.50 )
=
9.12  7.86
= 1.52
Consider (8.66, 0.72),
0.72 = lg A + 1.5(8.66)
lg A = -12.27
 A = 5.4 x 10-13 yr km-1.52
(b) For Earth,  = 1 year
 lg  = 0
From graph,
lg r = 8.19
Hence, r = 1.54 x 108 km
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10 N99/III/2
a Explain how an object travelling in a circle with constant speed has an acceleration. In
what direction is this acceleration?
b
A satellite P of mass 2400 kg is placed in a geostationary orbit at a distance of
4.23 × 107 m from the centre of the Earth.
(i) Explain what is meant by the term geostationary orbit
(ii) Calculate
1. the angular velocity of the satellite,
2. the speed of the satellite,
3. the acceleration of the satellite,
4. the force of attraction between the Earth and the satellite,
5. the mass of the Earth.
c
Explain why a geostationary satellite
(i) must be placed vertically above the equator,
(ii) must move from west to east.
d
Why is a satellite in a geostationary orbit often used for telecommunication?
a
An object travelling in a circle has its velocity changing since it is constantly changing its
direction of motion. Velocity is a vector quantity that has both magnitude and direction.
Having its direction changed constantly, the object must be experiencing an acceleration
(which is rate of change of velocity). The direction of acceleration is towards the center
of the circular motion (thus it is named as centripetal acceleration).
b
(i) The term geostationary orbit refers to a circular orbit around the Earth on which a
satellite would appear stationary to an observer on the Earth's surface.
2
(ii) 1.  
T
2

24  60  60
= 7.27 × 10-5 rad s-1
2. v  r 
v  4.23  107  7.272  105
= 3080 m s-1
3. a  r  2
a  4.23  107  7.272  105 
2
= 0.224 m s-2
4. Fc  mac
Fc  2400  0.2237
= 537 N
Since force of attraction between Earth and satellite is the only force acting on
the satellite, FG = Fc = 537 N.
5. Force of attraction between Earth and satellite is also given by:
GMm
F
r2
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536.9 
6.67  1011  M  2400
 4.23  10 
7 2
M = 6.00 × 1024 kg
c
(i) A geostationary satellite must be placed vertically above the equator because
o the centripetal force acting on the satellite must lie in the orbiting plane of the
point that the satellite is placed above. Hence the satellite can be always orbiting
directly above a point on the equator.
o if the orbit is not on the equator, the satellite will be sometimes over the northern
hemisphere and sometimes over the southern hemisphere and so cannot be
geostationary.
(ii) A geostationary satellite must move from west to east because the Earth's surface
moves from west to east. In order to appear stationary to an observer on the Earth's
surface, it must move in the same direction as Earth.
(iii) A satellite in geostationary orbit is often used as telecommunication because it will
remain at the same relative location, in constant electronic line-of-sight, to the datareceiving stations on the Earth's surface.
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