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17.2Gravitational potential and energy
This section deals with the concepts of gravitational potential and gravitational potential energy that are needed to understand motion in a gravitational field.
Consider a mass M placed somewhere in space, and a second mass m that is a distance r from M. The two masses share gravitational potential energy, which is stored in their gravitational field. This energy is the work that was done in bringing the two masses to a distance r apart from an initial separation that was infinite. For all practical purposes we consider M to be fixed in space and so it is just the small body of mass m that is moved (Figure 17.11).
KEY POINT
The gravitational potential energy of two bodies is the work that was done in bringing the bodies to their present position from when they were infinitely far apart.
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Figure 17.11: Work is done to bring the small mass from infinity to a given position near the big mass. The red arrow is the force of attraction between the two masses. We are interested in the work done by the force represented by the green arrow, i.e. the work done by the external agent. This work is negative and is stored as potential energy. The work done is independent of the actual path followed.

Notice that, strictly speaking, this is energy that belongs to the pair of masses M and m and not just to one of them. Notice also that when we say that the masses are moved from infinity we are implying that they are moved at a very small constant speed, so that no kinetic energy is involved.
The gravitational force is not constant, so we cannot straightforwardly calculate the work done. We need calculus for this calculation. The total work done in moving the mass m from infinity to a distance r from the centre of a spherical mass M is:
W=â«râGMmr2Â dr=â[GMmr]râ=âGMmr
Note that this is the work done by an external agent in bringing the mass m from far away to a position near M at constant speed. The force this agent exerts on m is equal and opposite to the gravitational force exerted on m by M. It is important to be clear about who exerts forces on whom and who does work; otherwise things can get very confused.
So this work is now the gravitational potential energy of the two masses when their centres are separated by a distance r:
Ep=âGMmr
This energy is negative. There are two ways to understand why this energy is negative:
1Referring to Figure 17.11 we see that the force is directed to the right and motion to the left. Since cos 180Â° = â1, the work done, and hence the potential energy, will be negative.
2Imagine that m is very far from M. We expect no potential energy in this case. If m is free to move it will be attracted by M and so m will increase its kinetic energy. So the potential energy must decrease (that is, it will become negative).
This implies that if we have a mass m on the surface of a planet of radius R and we want to separate the two masses by moving m a large distance away we must provide an amount of energy equal to + GMmR to m (Figure 17.12).

Figure 17.12: The mass m at the surface of a planet of mass M has negative potential energy. It needs energy to âget outâ of the âwellâ (to move away from the planetâs surface).

WORKED EXAMPLE 17.6
Calculate the difference in the potential energy of a satellite of mass 1500 kg when it is taken from the surface of the earth (mass M = 6.0 Ã 1024 kg, radius R = 6.4 Ã 106 m) to a distance of 500 km above the earthâs surface.
Answer
The potential energy of the system initially is:
EP=âGMmR=â6.67Ã10â11Ã6.0Ã1024Ã15006.4Ã106ââ9.4Ã1010Â J
At a distance of 520 km from the earthâs surface the separation of the masses is r = R + 520 km. The potential energy is therefore:
Ep=âGMmr=â6.67Ã10â11Ã6.0Ã1024Ã15006.9Ã106ââ8.7Ã1010Â J
The difference in these potential energies is:
(â8.7Ã1010)â(â9.4Ã1010)=7.0Ã109Â J
This difference of 7.0 Ã 109 J is the energy that needs to be provided to move the satellite from earth to its new position.
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Gravitational potential
Related to the concept of gravitational potential energy is that of gravitational potential, Vg.
KEY POINT
The gravitational potential at a point P in a gravitational field is the work done per unit mass in bringing a small point mass from infinity to point P. The unit of potential is J kgâ1.
If the work done is W, then the gravitational potential is the ratio of the work done to the mass m, that is:
Vg=Wm
The gravitational potential is a scalar quantity. Since W=âGMmr, the potential a distance r from a spherical mass M is:
Vg=âGMm/rm
Vg=âGMr
In general, if mass M produces a gravitational potential Vg at some point and we place a mass m at that point, the gravitational potential energy of the system is EP = mVg.
If a mass m is positioned at point A in a gravitational field where the gravitational potential is VgA and is then moved to another point B where the gravitational potential is VgB, then the work that is done on the mass by the external agent moving the mass is the difference in gravitational potential energy:
W=ÎEP
W=mVgBâmVgA
W=mÎVg
The work done does not depend on the actual path taken (Figure 17.13).

Figure 17.13: Work must be done to move a mass from one point to another in a gravitational field.

EXAM TIP
This formula is used when an external agent does work to move a mass m from one point in a gravitational field to another at constant small speed. The work done by the gravitational force is the opposite.
CHECK YOURSELF 5
The gravitational potential at a point P near a planet is â 2.0 Ã 1010 J kgâ1. A body of mass 350 kg is placed at P. What is the gravitational potential energy of the body and the planet?
WORKED EXAMPLE 17.7
Look at Figure 17.14. The gravitational potential at A is VA = â5 Ã 1012 J kgâ1and at B it is VB = â2 Ã 1012 J kgâ1.
What is the work done in moving a mass of 200 kg from A to B by:
aan external agent (green arrow)
bthe gravitational force (red arrow)?

Figure 17.14

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Answer
The external agent has to apply a force to the right (green arrow) equal to the gravitational force (red arrow) exerted on the body by the large mass that created the gravitational potential. The forces have to be equal so that the body is moved from A to B at a small constant speed.
aThe work done by the external agent is
W=m(VBâVA)=200Ã(â2Ã1012â(â5Ã1012))=6Ã1014Â J..
bThe work done by the gravitational force is â 6 Ã 1014 J.
CHECK YOURSELF 6
In worked example 17.7, the force at A is 1.7 Ã 109 N and the displacement from A to B is 9.0 Ã 105 m.
Why does the product 1.7 Ã 109 Ã 9.0 Ã 105 = 1.5 Ã 1015 J not give the right answer for the work done in moving the mass from A to B?
WORKED EXAMPLE 17.8
Figure 17.15 shows two spherical bodies. Calculate the gravitational potential at point P.

Figure 17.15

Answer
Gravitational potential is a scalar quantity, so we find the potential created by each mass separately and then add.
Potential from left mass:
V1=âGMr
V1=â6.67Ã10â11Ã4.4Ã10123.5Ã105
V1=â8.385Ã10â4Â JÂ kgâ1
Potential from right mass:
V2=â6.67Ã10â11Ã6.2Ã10102.0Ã105
V2=â2.068Ã10â5Â JÂ kgâ1
The total potential is then the sum V=V1+V2=â8.6Ã10â4J kgâ1.
WORKED EXAMPLE 17.9
The mass of the moon is about 81 times smaller than that of the earth. The distance between the earth and the moon is about d = 3.8 Ã 108 m. The mass of the earth is 6.0 Ã 1024 kg.
aDetermine the distance from the centre of the earth of the point on the line joining the earth to the moon where the combined gravitational field strength of the earth and the moon is zero.
bCalculate the combined gravitational potential at that point.
cCalculate the potential energy when a 2500 kg probe is placed at that point.
Answer
aLet the point we are looking for be point P, and the distance we are looking for be x. Let the mass of the earth be M, so the mass of the moon is M81.
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Use the information to draw a diagram to show the situation, as in Figure 17.16.

Figure 17.16

The field due to the earth at point P is then gearthÂ =GMx2.
The field due to the moon at point P is then gmoonÂ =GM81(dâx)2.
The combined field is zero. The two fields are opposite in direction and must be equal in magnitude. Therefore:
gearthÂ x=âx=gmoonÂ GMx2=GM81(dâx)21x2=181(dâx)2âx2=81(dâx)29(dâx)=9d10=910Ã3.8Ã108=3.4Ã108Â m
bThe combined potential at this point (we add the individual potentials for the earth and the moon since potential is a scalar quantity) is:
Vg=âGMxâ(GM81)dâx
Vg=â6.67Ã10â11Ã6.0Ã10243.4Ã108â6.67Ã10â11Ã6.0Ã10240.4Ã108Ã81
Vg=â1.3Ã106Â JÂ kgâ1
cUse the equation EP = mVg.
Substituting the mass of the probe for the mass m and using the value for Vg from part b:
Ep=2500Ã(â1.3Ã106)
Ep=â3.3Ã109Â J
WORKED EXAMPLE 17.10
The graph in Figure 17.17 shows the variation of the gravitational potential V with distance r away from the centre of a dense compact planet of radius 2 Ã 106 m.
Use the graph to calculate the work required to move a probe of mass 3400 kg from the surface to a distance of 7.5 Ã 106 m from the centre of the planet.

Figure 17.17

Answer
The work required can be found from W = mâVg or from the completely equivalent W = âEP.
The change in gravitational potential âVg is given by:
ÎVg=VfinalÂ âVinitialÂ 
ÎVg=â7.0Ã109â(â26Ã109)
ÎVg=19Ã109Â JÂ kgâ1
So the work done is:
W=mÎVg
W=3400Ã19Ã109
W=6.5Ã1013Â J
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Field and potential
Consider two points very close to each other separated by a distance Îr and suppose that the gravitational potential difference between the two points is ÎVg. The work done in moving a mass m from one point to the other can be calculated in two ways: mÎVg and FÎr. But F = mg and so mgÎr = mÎVg. This says that the magnitude of the gravitational field strength is:
g=ÎVgÎr.
This also says that g is the gradient of a graph of Vg versus r. (If calculus is used the correct expression is g=âdVgdr.)
CHECK YOURSELF 7
The graph in Figure 17.18 shows the variation of the gravitational potential along a line joining two planets.

Figure 17.18: For check yourself question 7.

What is g at point P?
Equipotential surfaces
We know that a spherical uniform mass M is the source of gravitational potential which, at a distance r from the centre of the source, is given by
Vg=âGMr
This means that points which are at the same distance from the centre of the mass are at the same potential. Points that have the same distance from the source lie on spheres. A two-dimensional representation of these spheres of constant potential is given in Figure 17.19. They are called equipotential surfaces. Notice that the field lines are normal to the equipotential surfaces.
An equipotential surface consists of those points that have the same potential.

Figure 17.19: Equipotential surfaces due to one spherical mass. These surfaces are drawn so that the difference in potential between any two adjacent surfaces is the same.

Figure 17.20a shows the equipotential surfaces of two equal spherical masses, and Figure 17.20b shows the surfaces for two unequal masses.

Figure 17.20: Equipotential surfaces for a two equal spherical masses and b two unequal masses. Field lines are normal to the equipotential surfaces.

The work done to move a point mass from one point to another point on the same equipotential surface is zero since W = mÎVg and ÎVg = 0 (Figure 17.21).

Figure 17.21: Zero work is done when moving a mass from one point to another on the same equipotential.

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TEST YOUR UNDERSTANDING
19Consider two particles of mass m and 16m separated by a distance d.
aDeduce that at point P, a distance d5 from the particle with mass m, the gravitational field strength is zero.
bDetermine the value of the gravitational potential at P.
20aWhat is the gravitational potential at a distance from the earthâs centre equal to 5 earth radii? (earth mass = 6.0 Ã 1024 kg, earth radius = 6.4 Ã 103 km.)
bWhat is the gravitational potential energy of a 500 kg satellite placed at a distance from the earthâs centre equal to 5 earth radii?
21aWhat is the gravitational potential energy stored in the gravitational field between the earth and the moon? (earth mass = 6.0 Ã 1024 kg, moon mass = 7.3 Ã 1022 kg, separation = 3.8 Ã 108 m)
bWhat is the earthâs gravitational potential at the position of the moon?
cFind the speed with which the moon orbits the earth.
22A spacecraft of mass 30 000 kg leaves the earth on its way to the moon and lands on the moon. Plot the spacecraftâs potential energy as a function of its distance from the earthâs centre. (The earthâmoon distance is 3.8 Ã 108 m. Take the mass of the earth to be 6.0 Ã 1024 kg and the mass of the moon to be 7.3 Ã 1022 kg.)
23Figure 17.22 shows the variation with distance from the centre of the planet of the gravitational potential due to the planet and its moon. The planetâs centre is at r = 0, and the centre of the moon is at r = 1. The units of separation are arbitrary. At the point where r = 0.75 the gravitational field is zero.

Figure 17.22: For question 23.

aDetermine the ratio of the mass of the planet to that of the moon.
bWith what speed must a probe be launched from the surface of the planet in order to arrive on the surface of the moon?
24Figure 17.23 shows a planet orbiting the sun counterclockwise at two positionsâA and B. Also shown is the gravitational force acting on the planet at each position.

Figure 17.23: For question 24.

By decomposing the force into components normal and tangential to the path (dotted lines), explain why it is only the tangential component that does work. Hence explain why the planet will accelerate from A to P but will slow down from P to B. Reach the same conclusions by energy considerations.
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25Figure 17.24 shows the variation of the gravitational force with distance between two masses.

Figure 17.24: For question 25.

What does the shaded area represent?
26Figure 17.25 shows equipotential surfaces for two spherical masses.

Figure 17.25: For question 26.

aExplain why the masses are unequal.
bDraw a few field lines for this arrangement.
cWhy are the surfaces further out becoming spherical?
dThe same mass is moved along two paths as shown. Compare the work done in each case.
