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In an uniform gravitational field the loss in potential energy (the work done) by gravity is given by $\text{-}mg\Delta h$ and the time to do so (assuming no friction and no initial velocity) is given my $\sqrt{\frac{2 \Delta h}{a}}$, where $a$ is the strength of the field as commonly referred to the local gravity.

Now, you can do the same for a non-uniform field. To get the work done you simply take the difference in gravitational potential energy: $\int_\infty^{h_1} G\frac{Mm}{{h_1}^2} - \int_\infty^{h_2} G\frac{Mm}{{h_2}^2}$.

And here is my problem. I'm stuck. How does one calculate the time it takes to fall from $h_1$ to $h_2$ with respect to the changing field strength?

Madde Anerson
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1 Answers1

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In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific energy implies that $$\mu\left(\frac{1}{r}-\frac{1}{R}\right) = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\eta}\frac{\mathrm{d}\eta}{\mathrm{d}t}\right)^2\text{,}$$ so basic substitution and some algebraic gymnastics can give you a differential equation for $\mathrm{d}t/\mathrm{d}\eta$, which has a solution $$t = \sqrt{\frac{R^3}{8\mu}}\left(\eta+\sin\eta\right)\text{.}$$ As a side note, the shape of this solution in an $(r,t)$-plane is a cycloid.

As James Cowley notes, there is a similarity to Kepler's third law. Specifically, a 'full orbit' would be from $\eta = 0$ to $\eta = 4\pi$, the second time $r = R$ is reached (the first time is just the other side of the orbit), which gives us $$t^2 = \frac{2\pi^2}{\mu}R^3\text{,}$$ as required by Kepler's third law, since this radial orbit is a limiting case of a high-eccentricity elliptic orbits.

Stan Liou
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