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17.3Motion in a gravitational field
Orbital motion
A satellite of mass m orbits a planet of mass M with speed v. The radius of the orbit is r (Figure 17.26). The total energy ET of this system is the sum of the kinetic energy EK and the gravitational potential energy EP.

Figure 17.26: A system of a satellite orbiting a planet.

EK=12mv2 and Ep=âGMmr
So:
ET=12mv2âGMmr
Note that we do not include any kinetic energy for the planet, as we assume it does not move.
Orbital speed is given by:
vorbitÂ =GMrâââââ
This means that the kinetic energy EK is:
EK=GMm2r
The total energy of the system becomes:
ET=GMm2râGMmr=âGMm2r
The total energy of an orbiting satellite is ET=âGMm2r.
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EXAM TIP
It is very important that you know how to derive the formula for orbital speed.
Figure 17.27 shows the kinetic energy EK, potential energy EP and total energy ET of a mass of 1 kg in orbit around the earth as a function of distance from the earthâs centre. This distance is measured in terms of the earthâs radius R.

Figure 17.27: Graphs of the kinetic, potential and total energy of a mass of 1 kg in circular orbit around the earth.

The law of gravitation combined with Newtonâs second law of motion allows an understanding of the motion of planets around the sun, as well as the motion of satellites around the earth. Suppose you launched an object from the surface of a planet with some speed. What would be the path followed by this object? Newtonâs laws give several possibilities.
âIf the total energy is positive, the object will follow a hyperbolic path and never return.
âIf the total energy is zero, the object will follow a parabolic path to infinity, where it will just about stop. It will never return.
âIf the total energy is negative, the object will go into a circular or elliptical orbit (or crash into the planet if the launching speed is too low).
Figure 17.28 illustrates these possible paths.

Figure 17.28: Launching a body from the surface of a planet results in various orbits, depending on the total energy ET of the body.

WORKED EXAMPLE 17.11
Evaluate the speed of a satellite in orbit at a height of 500 km above the earthâs surface and a satellite that just grazes the surface of the earth. (Take the radius of the earth to be 6.4 Ã 106 m and its mass 6.0 Ã 1024 kg.)
Answer
The speed is given by:
v2=GMr
The radius of orbit r of the satellite at a height of 500 km above the earthâs surface is:
r=(6.4Ã106)+(0.5Ã106)=6.9Ã106Â m
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Substituting for r and using the values from the question:
v=6.67Ã10â11Ã6.0Ã10246.9Ã106âââââââââââââââââââââ
âv=7.6Ã103Â m sâ1
For a grazing orbit, r = 6.4 Ã 106 m.
Using this value and following the same method, v = 7.9 Ã 103 m sâ1.
WORKED EXAMPLE 17.12
A satellite in a low orbit will experience a small frictional force (due to the atmosphere) in a direction opposite to the satelliteâs velocity.
aExplain why the satellite will move into an orbit closer to the earthâs surface.
bDeduce that the speed of the satellite will increase.
cDiscuss the fate of such a satellite.
Answer
aSince there is a frictional force acting, the satelliteâs total energy will be reduced.
The total energy ET of a satellite of mass m in a circular orbit of radius r around the earth of mass M is:
ET=âGMm2r
The masses do not change and G is a constant. The energy ET is negative, so reducing the energy means it becomes more negative. This means there must be a smaller radius, i.e. the satellite comes closer to the earth by spiraling inwards.
bThe speed of the satellite in a circular orbit is given by:
v=GMrâââââ
So we see that, as the satellite comes closer to earth, its speed increases.
cAs the satellite gets closer to earth it experiences more friction because the atmosphere is getting denser and the satellite moves faster. So as more and more energy is converted into thermal energy, the satellite most likely burns.
WORKED EXAMPLE 17.13
A probe of mass m is launched from the surface of a planet of mass M and radius R with kinetic energy EK=4GMm5R.
aExplain why this probe will not escape the gravitational field of the planet.
bThe probe eventually settles into a circular orbit around the planet with negligible work done by the engines. Calculate the radius of its orbit in terms of R.
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Answer
aThe total energy at launch is the sum of the kinetic and potential energies:
ET=4GMm5RâGMmR
ET=âGMm5R
This is negative and so the probe cannot escape.
bThe total energy in orbit is:
ET=âGMm2r
By energy conservation, the total energy once the probe is in orbit must equal the total energy at launch. So:
âGMm2r12r2rSoÂ r=âGMm5R=15R=5R=5R2
WORKED EXAMPLE 17.14
Figure 17.29 shows the variation of the gravitational potential due to a planet and its moon with distance r from the centre of the planet. The centre-to-centre distance between the planet and the moon is d. The planetâs centre is at r = 0, and the centre of the moon is at r = d.

Figure 17.29

State and explain the minimum energy required so that a 850 kg probe at rest on the planetâs surface will arrive on the moon.
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Answer
The probe will arrive at the moon provided it has enough energy to get to the peak of the curve. Once there, the moon will pull it in.
On the surface of the planet Vg = â3.9 Ã 108 J kgâ1. At the peak the potential is Vg = â0.4 Ã 108 J kgâ1. The minimum energy required is equal to the work done to move through this potential difference.
W=mÎVg
W=850Ã(â0.4Ã108+3.9Ã108)
W=3.0Ã1011Â J
So the minimum energy required is 3.0 Ã 1011 J.
Escape speed
Suppose that a body of mass m finds itself somewhere in the gravitational field of a planet or star. The object could be in orbit or just passing by or just placed there so that it is momentarily at rest. What is the minimum speed we should give to that body so that it moves far from the planet or star without returning? This is called the escape speed (Figure 17.30).
KEY POINT
Escape speed is the minimum speed a body in a gravitational field must have in order to reach infinity.

Figure 17.30: The escape speed will take the projectile from a point in the gravitational field to infinity, where it will come to rest.

When the body gets to infinity (or, at least, extremely far away) the gravitational potential energy is zero. Since escape speed is the minimum speed needed to escape, the speed at infinity will be zero. Hence the total energy will be zero. So the condition to escape from any point in a gravitational field is that the total energy at that point must be zero (and obviously you have to be moving away from the planet).
Consider an object momentarily at rest at a distance r from the centre of a planet of mass M.
The total energy is just gravitational potential energy and so equals ET=âGMmr.
So, we need to provide kinetic energy to make the total energy zero. In other words, the kinetic energy needed is:
EK=+GMmr
From 12mv2esc=GMmr we find vesc=2GMrââââââ.
Note that the escape speed is independent of the mass of the body escaping.
If the body is initially on the surface of a planet of radius R then the formula becomes vescÂ =2GMRââââââ.
The numerical value of this escape speed for the earth is about 11.2 km sâ1.
In practice, in order to escape from earth, a body must overcome not only the pull of the earth and the friction with the atmosphere but also the pull of the sun and the other planets. This means that the escape speed from the earth is larger than 11.2 km sâ1. This discussion does not apply to powered objects such as rockets; it applies only to objects launched from the earth like old fashioned cannon balls. In other words, it applies to ballistic motion only.
Consider now an object that in a circular orbit of radius r. What additional speed do we need to give the body in order that it escapes?
The escape speed is vescÂ =2GMrââââââ but the body already has an orbital speed vorbÂ =GMrâââââ and so the additional speed needed is 2GMrâââââââGMrâââââ.
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THEORY OF KNOWLEDGE
Changing ideas
The motion of the planets is the perfect application of the theory of gravitation. The understanding of the motion of the planets has undergone very many âparadigm shiftsâ since ancient times. Newton was motivated by the âlawsâ discovered by Kepler. Keplerâs laws were published in 1619 in a book called the Harmony of the World, nearly 70 years before Newton published his work. In ancient times, Ptolemy constructed an involved system in which the sun and the planets orbited the earth. The Ptolemaic worldview prevailed for centuries until Copernicus, early in the sixteenth century, asserted that the sun was at the centre of the motion of the planets in the solar system. Newtonâs law of gravitation has had great success in dealing with planetary motion but cannot account for some small irregularities, such as the precession of the orbit of Mercury and the bending of light near very massive bodies. In 1915, Einstein introduced the general theory of relativity, which replaced Newtonâs theory of gravity and resolved the difficulties of the Newtonian theory.
WORKED EXAMPLE 17.15
A probe is launched from the surface of a planet of mass M and radius R with a speed that is half the escape speed.
aCalculate, in terms of R, the largest distance from the surface of the planet the probe will get to.
bOnce the probe gets to the point in part a, rockets are fired giving the probe a speed v. What should v be so that the probe escapes?
Answer
aLet r be the largest distance from the centre of the planet reached by the probe.
The total energy of the probe at that distance is ET=âGMmr, since the kinetic energy is zero.
At launch the total energy is ET=12mv2âGMmR.
Now use the fact that v=vescÂ 2 to find the total energy at launch:
ET=12mv2escÂ 4âGMmR=12m(2GMR)4âGMmR=GMm4RâGMmR=â34GMmR
By energy conservation, the total energy at launch is equal to the total energy at the farthest distance, so:
â34GMmR34Râr=âGMmr=1r=4R3
The height from the surface is therefore h=râR=R3, a perhaps surprising result.
bUsing our formulae, the speed must be:
vesc=2GMrââââââ=2GM(4R3)âââââââîâ·îîî=3GM2Rââââââ.
WORKED EXAMPLE 17.16
aWhat must the radius of a star of mass M be such that the escape speed from the star is equal to the speed of light, c?
bCalculate this radius for the sun and the earth. (Use a mass of 2 Ã 1030 kg for the sun and 6 Ã 1024 kg for the earth.)
Answer
aUsing vescÂ =2GMRââââââ with vescÂ =c, we find R=2GMc2.
Since nothing can exceed the speed of light, if the radius of the star is equal to or less than 2GMc2, nothing can escape from the star. It is a black hole.
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bFor the sun:
R=2GMc2
R=2Ã6.67Ã10â11Ã2Ã1030(3Ã108)2
Râ3Ã103Â m
A similar calculation for the earth gives R = 9 mm. This shows that both the earth and the sun are far from being black holes!
Weightlessness
Consider an astronaut of mass m in a spacecraft in orbit around the earth a distance r from the earthâs centre (Figure 17.31). Why does the astronaut âfeel weightlessâ? At a distance of 300 km from the earthâs surface, gravity is by no means negligible. A simple answer is that the astronaut, as well as the spacecraft, are both falling freely, i.e. they have the same acceleration. Hence there are no reaction forces from the floor.

Figure 17.31: An astronaut inside a spacecraft or out in a spacewalk accelerates towards the earth with the same acceleration as the spacecraft. So whether inside or outside, there is no force of reaction on the astronaut. They feel âweightlessâ.

Quantitatively, the forces on the astronaut are the reaction force N from the floor of the spacecraft and his or her weight W (i.e. the gravitational force from the earth). The net force is
F=WâN
The astronautâs weight is given by:
W=GMmr2
So the net force F on the astronaut is:
F=GMmr2âN
This is the force that provides the acceleration keeping the astronaut in orbit at radius r, so:
GMmr2âN=mv2r
âN=GMmr2âmv2r
N=mr(GMrâv2)
But the orbital speed of the astronaut is given by:
v2orbitÂ =GMr
Substituting this into the equation for N gives N = 0. Thus, the astronaut experiences no reaction forces from the floor and so âfeelsâ weightless.
NATURE OF SCIENCE
The Newtonian theory of gravity and mechanics applied to the motion of planets gives spectacular agreement with the observed motion of the planets. However, during the nineteenth century very small discrepancies between theory and observations began to arise. One of them had to do with the motion of the planet Uranus. What could be the cause of these discrepancies? One possibility was that Newtonian gravity was not correctâan unlikely possibility, given the previous successes of the theory. The French astronomer Urbain Le Verrier (1811â1877) argued that a new, unknown planet was affecting the orbit of Uranus and was responsible for the erratic behaviour of its orbit. Working backwards, Le Verrier calculated the possible orbit of the unknown planet that would give rise to the observed effects on Uranus. He communicated his results to the Berlin observatory. His letter arrived at the observatory on 23 September 1846, and on that same evening the planet Neptune was discovered!
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TEST YOUR UNDERSTANDING
27Figure 17.32 shows cross-sections of two satellite orbits around the earth. (To be in orbit means that only gravity is acting on the satellite.) Discuss whether either of these orbits is possible.

Figure 17.32: For question 27.

28In the text it was calculated that the acceleration due to gravity at a height of 300 km above the earthâs surface is far from negligible, yet astronauts orbiting in a space shuttle at such a height feel weightless. Explain why.
29A rocket is launched from the surface of a planet. At the position shown in Figure 17.33, the rocket is a distance of 2R from the centre of the planet (where R is the radius of the planet) and its speed is v=GM2Râââââ. At that point the fuel runs out.

Figure 17.33: For question 29.

aExplain why the rocket will eventually crash onto the surface of the planet.
bCalculate, in terms of R, the maximum distance from the centre of the planet the rocket travels to.
cDetermine the speed with which the rocket crashes onto the planet surface.
dDraw a graph to show how the speed of the rocket varies with distance r from the centre of the planet as the rocket begins to fall back towards the planet.
30Prove that the total energy of the earth (mass m) as it orbits the sun (mass M) can be expressed as either E=â12mv2 or E=âGMm2r
where r is the radius of the earthâs circular orbit.
31A satellite orbits the earth in a circular orbit. A small frictional force acts on the satellite. State and explain the effect of this force on the following properties of the satellite:
athe total energy
bthe orbit radius
cthe speed
dthe potential energy
ethe period.
32Figure 17.34 shows two identical satellites in circular orbits.

Figure 17.34: For question 32.

Which satellite has the larger:
akinetic energy
bpotential energy
ctotal energy?
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33The total energy of a satellite during launch from the earthâs surface is E=âGMm5R, where R is the radius of the earth.
aExplain why this satellite will not escape the earth.
bThe satellite eventually settles into a circular orbit. Calculate the radius of the orbit in terms of R.
34A satellite is in a circular orbit around the earth. The satellite turns on its engines and now finds itself in a new circular orbit of larger radius. State and explain whether the work done by the engines is positive, zero or negative.
35Figure 17.35 shows a planet orbiting the sun.

Figure 17.35: For question 35.

aExplain why at points A and P of the orbit the potential energy of the planet assumes its minimum and maximum values, and determine which is which.
bHence determine at what point in the orbit the planet has the greatest speed.
cExplain why the angular momentum of the planet is constant.
dHence determine the ratio of the speeds at P and A in terms of the distances of P and A to the sun.
36Referring to the figure of the previous problem, show that the rate at which the planet sweeps area is given by ÎAÎt=12r2Ï where Ï is the angular speed of the planet when at distance r from the sun. Hence deduce that ÎAÎt=L2m where L is the angular momentum of the planet and m is its mass, and explain Keplerâs second law.
37Show that the escape speed from the surface of a planet of radius R can be written as vescÂ =2gRâââââ and vescÂ =â2Vâââââ where g is the gravitational field strength and V the gravitational potential at the planetâs surface. Show that yet another expression for the escape speed is vescÂ =R8ÏGÏ3ââââââ where Ï is the density of the planet.
38aDeduce that a satellite orbiting a planet of mass M in a circular orbit of radius r has a period of revolution given by T=4Ï2r3GMâââââââ.
bA grazing orbit is one in which the orbit radius is approximately equal to the radius R of the planet. Deduce that the period of revolution in a grazing orbit is given by T=3ÏGÏââââ where Ï is the density of the planet.
cThe period of a grazing orbit around the earth is 85 minutes, and around the planet Jupiter it is 169 minutes. Deduce the ratio ÏearthÂ ÏJupiterÂ .
39aThe acceleration of free fall at the surface of a planet is g, and the radius of the planet is R. Deduce that the period of a satellite in a very low orbit is given by T=2ÏRgâââ
bGiven that g = 4.5 m sâ2 and R = 3.4 Ã 106 m, deduce that the orbital period of the low orbit is about 91 minutes.
cA spacecraft in orbit around this planet has a period of 140 minutes. Deduce the height of the spacecraft from the surface of the planet.
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40Two stars of equal mass M orbit a common centre as shown in Figure 17.36. The radius of the orbit of each star is R. Assume that each of the stars has a mass equal to 1.5 solar masses (solar mass = 2.0 Ã 1030 kg) and that the initial separation of the stars is 2.0 Ã 109 m.

Figure 17.36: For question 40.

aState the magnitude of the force on each star in terms of M, R and G.
bDeduce that the period of revolution of each star is given by the expression:
T2=16Ï2R3GM
cEvaluate the period numerically.
dShow that the total energy of the two stars is given by:
E=âGM24R
eThe two-star system loses energy as a result of emitting gravitational radiation. Deduce that the stars will move closer to each other.
fiExplain why the fractional loss of energy (ÎEE) per unit time may be calculated from the expression:
ÎEEÎt=2ÎTT3Ît
where ÎTTÎt is the fractional decrease in period per unit time. (Hint: use ideas of error propagation.)
iiThe orbital period decreases at a rate of ÎTÎt=72Â Î¼sgrâ1. Estimate the fractional energy loss per year.
gThe two stars will collapse into each other when ÎE â E. Estimate the lifetime, in years, of this binary star system.
