Weighing the Earth

Before beginning to make use of the law of universal gravitation, we must turn our attention to an important detail.

We have just calculated the force of attraction between two loads located at a distance of from each other. But if this distance were ? What would we then substitute in the formula—the distance between the surfaces of bodies or the distance between their centres of gravity or some other value?

The law of universal gravitation, , can be applied with complete rigour when such doubts do not arise. The distance between the bodies should be much greater than their dimensions; we should have the right to regard the bodies as points. But how should we apply the law to two nearby bodies? This is simple in principle: we must conceptually break up the bodies into small pieces, calculate the force for each pair and then add (vectorially) all the forces.

In principle this is simple, but it is rather complicated in practice.

However, nature has helped us. Computations show that if the particles of a body interact with a force proportional to , spherical bodies possess the property of attracting like points located at the centres of the spheres. For two nearby spheres, the formula is exactly valid, just as for distant spheres, if is the distance between their centres. We have already used this rule above in computing the acceleration on the Earth’s surface.

We now have the right to apply the gravitational formula for computing the forces with which the Earth attracts bodies. We should take the distance from the centre of the Earth to the body as . Let be the mass, and the radius of the Earth. Then the force of attraction acting on a body of mass at the Earth’s surface But this is in fact the body’s weight, which we always express as . Hence, the acceleration of free fall Now at last we can say how the Earth was weighed. The quantities , and are known, so the Earth’s mass can be computed from this formula. The Sun can also be weighed in the same manner.

But can we really call such a procedure weighing? Of course we can; indirect measurements play at least as great a role in physics as direct measurements.

Let us now solve a curious problem.

An essential role in the plans for creating world-wide television is played by the creation of a “24-hour satellite”, i.e. one which will always be situated over one point on the Earth’s surface. Will such a satellite experience a significant frictional force? This depends on how far from the Earth it will have to perform its rotation.

A 24-hour satellite should revolve with a period equal to 24 hours. If is the distance from the satellite to the centre of the Earth, then its speed and its acceleration On the other hand, this acceleration whose source is the Earth’s attraction is equal to Equating our two expressions for the acceleration, we obtain: Substituting the founded-off values of , and , we obtain: There is no air friction at such a height, and a 24-hour satellite will not slow down its “motionless orbiting”.