University Physics I - Classical Mechanics, 2019

10.1. THE INVERSE-SQUARE LAW 223
near the surface of any other planet or moon, just replacing M E and R E by the mass and radius of
the planet or moon in question. Thus, we could speak of g moon , g Mars , etc., and in some homework
problems you will be asked to calculate these quantities. Clearly, besides telling you how fast things
fall on a given planet, the quantity g planet allows you to figure out how much something would weigh
on that planet’s surface (just multiply g planet by the mass of the object); alternatively, the ratio of
g planet to g E will be the ratio of the object’s weight on that planet to its weight here on Earth.
Of course, historically, this is not what Newton and his contemporaries would have done: they had
measurements of objects in free fall (or sliding on inclined planes) that would have given them the
value of g, and they even had a pretty good idea of the radius of the Earth 1 , but they did not
know either G or the mass of the Earth, so all they could get from Eq. (10.3) was the value of the
product GM E . It was only a century later, when Cavendish measured G, that they could get from
that the mass of the earth (as a result of which, he became known as “the man who weighed the
earth”!)
What Newton could do, however, with just this knowledge of the value of GM E ,wassomething
that, for its time, was even more dramatic and far-reaching: namely, he could “prove” his intuition
that the same fundamental interaction—gravity—that causes an apple to fall near the surface of
the Earth, reaching out hundreds of thousands of miles away into space, also provides the force
needed to keep the moon on its orbit. This brought together Earth-bound science (physics) and
“celestial” science (astronomy) in a way that no one had ever dreamed of before.
To see how this works, let us start by assuming that the moon does move in a circle, with the Earth
motionless at the center (these are all approximations, as we shall see later, but they give the right
order of magnitude at the end, which is all that Newton could have hoped for anyway). The force
F G E,m then has to provide the centripetal force F c = M moon ω 2 r e,m ,whereω is the moon’s angular
velocity. We can cancel the moon’s mass from both terms and write this condition as
GM E
r 2 e,m
= ω 2 r e,m (10.4)
The moon revolves around the earth once about every 29 days, which is about 29 × 24 × 3600 =
2.5 × 10 6 s. So ω is 2π radians per 2.5 million seconds, or ω =2.5 × 10 −6 rad/s. Substituting this
into Eq. (10.4), as well as the result GM E = gRE 2 (note that, as stated above, we do not need to
know G and M E separately), we get r e,m =3.99 × 10 8 m, pretty close to today’s accepted value of
the average Earth-moon distance, which is 3.84 × 10 8 km. Newton would not have known r e,m to
such an accuracy, but he would still have had a pretty good idea that this was, indeed, the correct
order of magnitude 2 .
1 The radius of the earth was known, at least approximately, since ancient times. The Greeks first figured it out
by watching the way tall objects disappear below the horizon when viewed from a ship at sea. Next time somebody
tries to tell you that people used to believe the earth was flat, ask them just which “people” they are talking about!
2 And, yes, the first estimates of the distance between the earth and the moon also go back to the ancient Greeks!
You can figure it out, if you know the radius of the earth, by looking at how long it takes the moon to transit across
the earth’s shadow during a lunar eclipse.