Gravity Distance Laws – Insights from ‘A Brief History of Time’

distancegravitynewtonian-gravityorbital-motion

In “A Brief History of Time”, Hawking explains Newtonian gravity in Chapter 2, Space and Time. The further apart the bodies are, the smaller the force. The gravitational attraction of a star is exactly one quarter that of a similar star at half the distance. This is because as we know, gravitational force is inversely proportional to the square of the distance between the bodies.

Next comes the part that confuses me.

If the law were that the gravitational attraction of a star went down faster with distance, the orbits of the planets would not be elliptical, they would spiral in to the sun. If it went down slower, the gravitational forces from the distant stars would dominate over that from the earth.

I am unclear as to what the meaning of "went down faster with distance". Gravitational force is indeed going down with distance. By adding "faster", does he mean the rate of change of gravity with distance, which would then be the derivative of $r^{-2}$ ($r$ is distance) = $-2*r^{-3}$. This tells us the gravity goes down faster with distance? I am confused. Can someone explain what he is trying to say and how we can conclude that this would either cause the planets to spiral in to the sun or how distant stars would dominate over that from the earth.

Best Answer

As you say in your question the gravitational force is proportional to $1/r^2$ so we can write it as:

$$ F \propto \frac{1}{r^n} \tag{1} $$

where $n = 2$. When Hawking talks about gravity decreasing faster with distance he means that in equation (1) $n > 2$, and if gravity decreases more slowly with distance that means $n < 2$. The value of $n$ is (hypothetically at least) not necessarily an integer.

Hawking is referring to Bertrand's theorem, which tells us that we have stable closed orbits only when $n = 2$.

As it happens the force is not quite an inverse square law. When we do a more accurate calculation using General Relativity we find the force is very slightly different from a $1/r^2$ force. The difference is tiny, but measurable, as it leads to the anomalous precession of Mercury.

Related Question