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17.1Newtonâs law of gravitation
Newtonâs second law implies that, whenever a particle moves with acceleration, a net force must be acting on it.
You may have heard the story about Newton sitting under an apple tree. A falling apple made him think about the nature of gravitation.
The apple, falling freely under gravity, would have been accelerating at 9.8 m sâ2. It would have experienced a net force in the direction of the acceleration. This force is what we call the âweightâ of the apple.
Similarly, a planet that orbits around the sun also experiences acceleration, and thus a force is acting on it.
Newton hypothesised that the force responsible for the falling apple is the same kind of force as that acting on a planet as it moves around the sun.
KEY POINT
Newton proposed that the attractive force of gravitation between two point masses (idealised bodies with no physical size) is given by the formula:
F=GM1M2r2
where M1 and M2 are the masses of the bodies, r is the distance between them and G is a constant. It has the value G = 6.67 Ã 10â11 N m2 kgâ2. The direction of the force is along the line joining the two masses.
This formula applies to point masses, but it can also be applied to any two bodies provided the distance between them is very large compared to the physical size of the bodies. In addition, Newton proved that for bodies that are spherical and of uniform density, one can assume that the entire mass of the body is concentrated at its centreâso in this case, the body acts as if it were a point mass. The distance r in the formula is the centre-to-centre separation of the spherical bodies.
THEORY OF KNOWLEDGE
Deterministic system
The laws of mechanics, along with Newtonâs law of gravitation, are the basis of classical physics. They describe a perfectly deterministic system. This means that if we know the positions and velocities of the particles in a system at some instant of time, then the future positions and velocities of the particles can be predicted with absolute certainty. Since the beginning of the twentieth century we have known that this is not true in many cases. In situations normally associated with âchaosâ the sensitivity of the system to the initial conditions is such that it is not possible to make accurate predictions of the future state.
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Figure 17.1 shows the gravitational force between two masses. The gravitational force is always attractive. The magnitude of the force on each mass is the same. This follows both from the formula as well as from Newtonâs third law.

Figure 17.1: The mass of the spherical body to the left can be thought to be concentrated at its centre.

CHECK YOURSELF 1
The distance between two point masses is tripled. What happens to the force between the two masses?
WORKED EXAMPLE 17.1
The average distance between the earth and the sun is 1.5 Ã 1011 m.
The mass of the earth is 6.0 Ã 1024 kg.
The mass of the sun is 2.0 Ã 1030 kg.
aEstimate the force between the sun and the earth.
bWhat is the acceleration experienced by earth?
Answer
aSubstituting these values into the formula F=GM1M2r2 gives:
F=(6.67Ã10â11)Ã(6.0Ã1024)Ã(2.0Ã1030)(1.5Ã1011)2=3.6Ã1022Â N
bÂ acceleration of the earthÂ =FME=3.6Ã10226.0Ã1024=6.0Ã10â3Â mÂ sâ2
Gravitational field strength
For hundreds of years, many physicists and philosophersâincluding Newtonâhave wondered how a mass âknowsâ about the presence of another mass nearby that will attract it. By the nineteenth century, physicists had developed the idea of a âfieldâ. This provided a (partial) answer to the question.
A mass M is said to create a gravitational field in the space around it. This means that, when another mass is placed at some distance from M, it âfeelsâ the gravitational field in the form of a gravitational force.
KEY POINT
The gravitational field strength, at a certain point, is the gravitational force per unit mass experienced by a small point mass m placed at that point:
g=Fm
The unit of gravitational field strength is N kgâ1 (equivalent to m sâ2). It is a vector quantity.
Turning this around, we find that the gravitational force on a point mass m is F = mg. But this is the expression we previously called the weight of the mass m.
So, we learn that the gravitational field strength is the same as the acceleration of free fall.
The force experienced by a small point mass m placed at distance r from a (spherical) mass M is:
F=GMmr2
So the gravitational field strength (Fm) of the spherical mass M is:
g=GMr2
THEORY OF KNOWLEDGE
Law of gravitation
Where did the law of gravitation come from? Not just from Newtonâs great intuition but also from the knowledge obtained earlier by Kepler, that planets move around the sun with a period that is proportional to the 32 power of the average orbit radius. To get such a law, the force of gravitation had to be an inverse square law.
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CHECK YOURSELF 2
The gravitational field strength on the surface of a planet is g. A satellite orbits the planet at a height from the surface equal to the radius of the planet.
What is the magnitude of the acceleration of the satellite?
The direction of the gravitational field strength at some point is given by the direction of the force a point mass would experience if placed at that point. The gravitational field strength around a single point or spherical mass is radial, which means that it points towards the centre of the mass creating the field. This is illustrated in Figure 17.2. This field is not uniformâthe field lines get farther apart with increasing distance from the point mass, showing that the magnitude of g decreases as we move away from M.

Figure 17.2: The gravitational field around a point (or spherical) mass is radial.

By contrast, Figure 17.3 shows a field with constant gravitational field strength. Here the field lines are equally spaced and parallel. The assumption of constant acceleration of free fall (which we used for projectile motion in ChaptersÂ 1â4) corresponds to this case.

Figure 17.3: The gravitational field above a flat mass is uniform.

CHECK YOURSELF 3
The gravitational field strength at a point P in a gravitational field is 15 N kgâ1. A point mass of 0.40 kg is placed at P.
What is the gravitational force on the mass?
WORKED EXAMPLE 17.2
The distance between two bodies is doubled.
Predict what will happen to the gravitational force between them.
Answer
Since the force is inversely proportional to the square of the separation, doubling the separation decreases the force by a factor of 22 = 4.
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WORKED EXAMPLE 17.3
Determine the acceleration of free fall (the gravitational field strength) on a planet 10 times as massive as the earth and with a radius 20 times as large.
Answer
From g=GMr2 we find:
g=G(10ME)(20rE)2
g=10GME400r2E
g=140GMEr2E
g=140gEâ0.25Â mÂ sâ2
EXAM TIP
For this type of problem write the formula for g and then replace mass and radius in terms of those for earth. It is a common mistake to forget to square the factor of 20 in the denominator.
WORKED EXAMPLE 17.4
Calculate the acceleration of free fall at a height of 300 km from the surface of the earth (the earthâs radius, RE, is 6.4 Ã 106 m and its mass is 6.0 Ã 1024 kg).
Answer
The acceleration of free fall is the same as the gravitational field strength. At height h from the surface:
g=GME(RE+h)2
where RE = 6.4 Ã 106 m is the radius of the earth. We can now put the numbers in:
g=6.67Ã10â11Ã6.0Ã1024(6.7Ã106)2
gâ9Nkgâ1
EXAM TIP
Notice the addition of the height to the radius of the earth.
Figure 17.4 shows two spherical masses M and m a distance d apart. We want to find the gravitational field strength at a point in between the two masses, on the line joining them (Figure 17.4).

Figure 17.4

At a point that is a distance x from the centre of the larger mass the net gravitational field strength is:
g=âGMx2+Gm(dâx)2
This is because the large mass creates a gravitational field GMx2 directed to the left and the small mass a gravitational field Gm(dâx)2 directed to the right.
A negative sign indicates a field pointing to the left and a positive sign a field pointing to the right.
Figure 17.5 shows the variation of g with xd. What is the ratio Mm?

Figure 17.5: The variation of g with distance from the large mass.

We see that at xd=0.75, g = 0.
This means:
âGM0.752+Gm0.252=0
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And so:
Mm=0.7520.252=9
Orbital motion
Figure 17.6 shows a particle of mass m orbiting a larger body of mass M in a circular orbit of radius r. To maintain a constant orbit there must be no frictional forces, so the only force on the particle is the force of gravitation, F=GMmr2.
This force provides the centripetal force on the particle. Therefore:
mv2r=GMmr2
Cancelling the mass m and a factor of r, this leads to:
v=GMrâââââ

Figure 17.6: A particle of mass m orbiting a larger body of mass M in a circular orbit of radius r.

This gives the speed in a circular orbit of radius r. Notice that the closer the satellite is to the surface, the higher the speed. But we know that v=2ÏrT, where T is the period of revolution (that is, the time to complete one full revolution). Squaring v=2ÏrT and equating the two expressions for v2, we deduce that:
4Ï2r2T2=GMr
âT2=4Ï2r3GM
This shows that the period of planets going around the sun is proportional to the 32 power of the orbit radius. Newton knew this from Keplerâs discoveries (see next section), so he knew that his choice of inverse distance squared in the law of gravitation was reasonable.
Keplerâs laws of planetary motion
Newton wrote down his law of gravitation in 1687. Between 1609 and 1619, the astronomer Johannes Kepler analysed the observations on the motion of planets made by the astronomer Tycho Brahe. Keplerâs spectacular achievement was to discover the three laws that govern the motion of the planets around the sun.
Keplerâs first law states that planets move on ellipses with the sun at one of the foci of the ellipse. (See Figure 17.7).

Figure 17.7: a The orbits of planets are elliptical. b The ellipse has two special points on its long axis, called the foci. The sum of the distances of any point on the ellipse from each of the foci is a constant.

The length of the long axis of the ellipse (the major axis) is called 2a. This length is important for Keplerâs third law.
Keplerâs second law states that the line joining a planet and the sun sweeps equal areas in equal times. (See Figure 17.8.)

Figure 17.8: a In equal times, the line joining the planet to the sun sweeps out equal areas. b The special case when the planet is at perihelion (P) and aphelion (A).

Figure 17.8b shows the planet at its closest (perihelion) and furthest (aphelion) from the sun. We approximate the areas swept by the planet when at P and A by the red triangles. In equal times the areas are the same. The area of the left triangle is 12Â baseÂ ÃÂ heightÂ =12vptÃSP. The right triangle area is similarly 12vAtÃSA. Equating the areas gives 12vPtÃSP=12vAtÃSA and so vPvA=SASP. This shows that the planet moves fastest at perihelion.
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Keplerâs third law states that the period of revolution of a planet around the sun is proportional to the 32 power of the semi-major axis of the ellipse (the length a in Figure 17.7). More precisely:
T2=4Ï2a3GM
For a circular orbit of radius r, r and a are the same.
Newton knew of Keplerâs discoveries. He realised that a force of attraction between two bodies that varied as F~1r2 would lead to elliptical or circular closed orbits (Keplerâs first law). He also knew that such a law would lead to Keplerâs third law. And finally by applying the law of conservation of angular momentum he could derive Keplerâs second law. The fact that Newtonâs laws lead to the empirical laws discovered by Kepler was solid evidence that Newton was right about his law of gravitation.
WORKED EXAMPLE 17.5
Halleyâs comet has a period of about 75 years. The perihelion distance (when it is closest to the sun) is 8.8 Ã 1010 m. Find the distance between the comet and the sun when the comet is at aphelion (furthest from the sun) and calculate the ratio of the perihelion to aphelion speed.
Answer
From T2=4Ï2a3GM we find:
a=(GMT24Ï2)13=ââââ6.67Ã10â11Ã2.0Ã1030Ã(75Ã365Ã24Ã3600)24Ï2ââ ââ13=2.7Ã1012Â m
Since aphelion distance + perihelion distance = 2a, it follows that the aphelion distance is 5.3 Ã 1012 m.
The perihelion to aphelion speed ratio is then 5.3Ã10128.8Ã1010â60.
CHECK YOURSELF 4
Why are comets such a rare sight?
Links
The prediction of the orbits of the planets was the crowning achievement of the Newtonian theory, so much so that soon after electrons were discovered, one of the first models of the atom was a miniature planetary system with electrons orbiting the nucleus of the atom in circular orbits. That was the Rutherford model which was soon replaced by the Bohr model which also had the electrons in circular orbits. So strong was the influence of the Newtonian planetary model that Sommerfeld modified the Bohr model to include elliptical electron orbits. But this reliance on the old ideas had to stop when new phenomena were observed that required a radically new approach to the whole subject.
NATURE OF SCIENCE
Predictions versus understanding
Combining the laws of mechanics with the law of gravitation enables scientists to predict with great accuracy the orbits of spacecraft, planets and comets. But to what degree do they enable an understanding of why planets, for example, move the way they do? In ancient times, Ptolemy was also able to predict the motion of planets with exceptional precision. In what sense is the Newtonian approach âbetterâ? Ptolemyâs approach was specific to planets and could not be generalised to other examples of motion, whereas the Newtonian approach can. Ptolemyâs method gives no explanation of the observed motions whereas Newton âexplainsâ the motion in terms of one single universal concept, that of a gravitational force that depends in a specific way on mass and separation. In this sense the Newtonian approach is superior and represents progress in science. But there are limits to the degree to which one demands âunderstandingâ: the obvious question for Newton would be, âWhy is there a force between two masses?â Newton could not answer this question.
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TEST YOUR UNDERSTANDING
Unless stated otherwise, take the radius and mass of the earth to be 6.4 Ã 106 m and 6.0 Ã 1024 kg, respectively.
1Calculate the gravitational force between:
athe earth and the moon (distance between earth and moon = 3.8 Ã 105 km; moonâs mass = 7.3 Ã 1022 kg)
ba proton and an electron separated by 10â10 m (mass of proton = 1.67 Ã 10â27 kg; mass of electron = 9.11 Ã 10â31 kg).
2A mass m is placed at the centre of a thin, hollow, spherical shell of mass M and radius r, as shown in Figure 17.9a.

Figure 17.9: For question 2.

aDetermine the gravitational force the mass m experiences.
bDetermine the gravitational force m exerts on M.
A second mass m is now placed a distance of 2r from the centre of the shell, as shown in Figure 17.9b.
cDetermine the gravitational force the mass inside the shell experiences.
dSuggest what gravitational force is experienced by the mass outside the shell.
3Stars A and B have the same mass, and the radius of star A is nine times larger than the radius of star B. Calculate the ratio of the gravitational field strength on star A to that on star B.
4Planet A has a mass that is twice as large as the mass of planet B and a radius that is twice as large as the radius of planet B. Calculate the ratio of the gravitational field strength on planet A to that on planet B.
5Stars A and B have the same density, and star A is 27 times more massive than star B. Calculate the ratio of the gravitational field strength on star A to that on star B.
6A star explodes and loses half its mass. Its radius becomes half as large. Determine the new gravitational field strength on the surface of the star in terms of the original one.
7The mass of the moon is about 81 times less than that of the earth. Estimate the fraction of the distance from the earth to the moon where the gravitational field strength is zero. (Take into account the earth and the moon only.)
8Figure 17.10 shows point P is halfway between the centres of two equal spherical masses that are separated by a distance of 2 Ã 109 m. Calculate the gravitational field strength at point P and state the direction of the gravitational field strength at point Q.

Figure 17.10: For question 8.

9Two satellites of mass m and 2m orbit the earth with the same orbit radius. Which satellite experiences the larger force from earth? Which satellite experiences the larger acceleration?
10Two satellites, X and Y, orbit the earth in circular orbits. The orbit radius of X is double that of Y. What is the ratio of the acceleration of X to that of Y?
11A satellite orbits the earth above the equator with a period equal to 24 hours.
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aDetermine the height of the satellite above the earthâs surface.
bSuggest an advantage of such a satellite.
12The Hubble Space Telescope is in orbit around the earth at a height of 560 km above the earthâs surface.
aCalculate Hubbleâs speed.
bIn a servicing mission, a space shuttle spotted the Hubble telescope a distance of 10 km ahead. Estimate how long it took the shuttle to catch up with Hubble, assuming that the shuttle was moving in a circular orbit just 500 m below Hubbleâs orbit.
13The solar system is at a distance of about 27,000 light years from the centre of the Milky Way galaxy. A light year is the distance travelled by light in one year. The solar system orbits the galactic centre in a circular orbit with a speed that is roughly 200 km sâ1.
aThe speed of light is 3.0 Ã 108 m sâ1. Convert a light year into metres.
bHow long, in years, does it take the solar system to perform one full revolution?
cThe mass of the galaxy that attracts our solar system is the mass of the galaxy that is within the orbit radius of the solar system. Estimate this mass.
14Which planet of the solar system has the greatest orbital speed? (Donât just look it up!)
15Assume that the force of gravity between two point masses is given by F=Gm1m2rn where n is a constant.
aDerive the law relating period to orbit radius for this force.
bDeduce the value of n if this law is to be identical to Keplerâs third law.
16aShow that a moon orbiting a planet of mass M in a circular orbit of radius r has a period T given by T2=4Ï2GMr3.
bEuropa is a moon of Jupiter with an orbit radius 6.7 Ã 105 km and period 3.6 days. Triton is a moon of Neptune with an orbital radius 3.5 Ã 105 km and period 5.9 days. What is the ratio of the mass of Jupiter to that of Neptune?
17The moon orbits the earth with an orbit radius 3.8 Ã 105 km and period 27 days. The radius of the earth is 6.4 Ã 103 km. From this information deduce the density of the earth.
18Tethys is a moon of Saturn with a period of 1.9 days and an orbit radius 2.9 Ã 105 km. Titan is another moon of Saturn with a period of 16 days. What is the orbit radius of Titan?
