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Universal Gravitation Applied

The first application of any new theory of physics is usually to explain previously unexplained experimental results. Newton did this be deriving Kepler's Laws of planetary motion. He immediately moved on to the next step, still followed today, predict measurements which have not or can not be made directly, but which can be verified by consistency checks with things that can be or have been measured. For Newton, one of the first checks along these lines was to determine the ratio of Jupiter's mass to the earth's mass. Even without the value of the gravitational constant, G, he could do this by application of his version of Kepler's Third Law.

1. Find the ratio of Jupiter's mass to the earth's mass:

   Solution: In 1610, Galileo and Simon Marius independently discovered Io, one of the moons of Jupiter. Subsequent observations showed that the period and mean orbital radius of Io were 42 hours and 421,600 km, respectively (you can get all kinds of astronomical information from http://dept.physics.upenn.edu/nineplanets/). Since both our moon and Io have to follow Keplerian orbits around earth and Jupiter, respectively, Kepler's Third Law gives
   

   MJupiter/Mearth	= 4*Pi2rIo3/(GTIo2) GTmoon2/(4*Pi2rmoon3)
   	= (rIo/rmoon)3 * (Tmoon/TIo)2
   	= (421,600 km/384,400 km)3 * [(2.36055 x 106 s)/(42 h x 3600 s/h)]2
   MJupiter/Mearth	= 322

   Modern values give the ratio as (1.900 x 10²7 kg)/(5.98 x 10²⁴ kg) = 318. Not bad for ol' Newton.

The Sphere problem

Another unsolved problem was Newton's extrapolation of a theory describing the gravitational force between point masses applied to objects which are clearly not points! For this reason, he developed differential calculus and showed its connection to integral calculus. Some arguments however do not explicitly need integral calculus, but rather a keen mathematical intuition. For example, Newton first showed that the gravitational force due to an empty spherical shell was the same as that due to a point at the center of the shell which contained all the mass of the shell. To do that, the easiest approach is to find the gravitational force due to a thin ring of mass M on a mass m₀ a distance x from the center of the ring as shown in the figure below.

The perspective might be difficult to see, but if we break the ring into infinitesimal lengths, dl, each with mass dm, then each of these masses exerts a force dF on m₀. All of these forces have the same magnitude since the distance r of each dm from m₀ is the same. Notice that, by symmetry, only the components of dF that lie along x will survive since all components perpendicular to x cancel (note the forces due to the pair of dm's shown in the figure where it should be clear that the perpendicular components are in opposite directions). So, to find the total force acting on m₀, we integrate up all of the components dFcos(theta) = dF(x/r). To do this we integrate over all the dm's since

Since x and r are constant, this integral just leads to the total force being

So the net force points towards the center of the ring.

Let's extend the result to a spherical shell of mass M as shown below and find the force due to the shell on a mass m₀ located a distance R from the center of the shell.

We break the shell into rings. If the shell has a radius of a, then the ring has a radius of asin(alpha) and a width adalpha. The ring has a mass

Again, only the component of the gravitational force which is parallel to the horizontal does not cancel when we consider the force due to the whole ring, so the gravitational force is

From the figure you should see that

Furthermore, the law of cosines gives

We wind up with an easier integral if we eliminate da, so lets differentiate our expression for r² to get

Putting this all back into our expression for dF, we get

We now integrate over all the circular strips. The limits of integration go from r = R - a to r = R + a.

Thus the force is the same as for a point mass M located at the center of the spherical shell.

There are three interesting bonuses to having suffered through this calculation. First, since every spherical shell has this form for the force, integrating over a bunch of shells forming a solid sphere does nothing to change the form of the force. Each shell acts like a point mass at the center so integrating over an infinite number of shells all stacked concentrically leads to the same result with the mass of the solid sphere replacing our value of M.

Another bonus is that the only limits on our derivation are that the density inside the spherical shell be constant (i.e. all the rings have the same mass). If the point mass m₀ is inside the spherical shell, we have only to change the limits of integration, the set up of the integral is unchanged. For R < a, The limits of integration are a - R to a + R. For these limits our integration yields

Hence the gravitational force inside a spherical shell is zero. We can also see this from a symmetry argument as shown in the figure below where we look at a point mass being attracted by some part of the mass of the shell on one side. The amount of mass is determined by the opening angle we decide to use, but if we use the same opening angle to estimate how much mass on the other side of the shell is also providing force on our mass inside the shell, we find the mass of both pieces of the shell go as r². However, since the force goes as 1/r², both pieces contribute the same amount of force, but in opposite directions so the net force is zero.

The last bonus is that we can extrapolate this result again to a solid sphere and note that any point inside a sphere has a gravitational force acting on it that depends only on the amount of mass which lies at a radius less than the points radius from the center. For example, for a sphere of mass M and radius R and a point of mass m at a radius r < R, the mass of material at radii less than r would be (assuming constant density as always)

where rho is the density. Since the distribution of M(r) is spherically symmetric, this still acts like a point mass at the center of the sphere as far calculating the gravitational force is concerned, so the force on the point at r is

Thus the force goes as r rather than 1/r².

Gravity as a Window Into the Earth and Beyond

Newton's success in understanding gravity, when coupled to Kepler's model of the solar system, led to a revolution in mankind's understanding of the universe and our confidence in being able to explain its mysteries. By carefully applying the ideas of bodies in orbit around one another, he not only explained the laws of planetary motion as discovered by Kepler, but correctly explained more down-to-earth phenomena such as the tides.

In the 20th century, we use Newton's Law of Gravitation to plan the path of space probes which give us close up views of other planets. The remarkable accuracy of a rather simple formulation of the working of gravity is only the least of Newton's achievement however. The Newtonian synthesis, the idea that laws of physics discovered on earth were the same laws operating in distant, unobserved parts of the universe, has been a guiding principle in our exploration of the universe.