Gravitational Force

Question

A thin disk of radius 'a' and mass M lies horizontally, a particle of mass m is at a height h directly above the center of the disk. The gravitational force F,exerted by the disk on the mass is given by $$F=({2GMmh\over a^2})({1\over h}-{1 \over (a^2+h^2)^{1/2}})$$ Assume $a<h$ and think of $F$ as a function of $a$, with the other quantities constant. Expand $F$ as a series in $a\over h$ showing only the first two nonzero terms.

Answer 1

As this is getting too long for comments, ...

$$F={2GMmh\over a^2}\cdot \left({1\over h}-{1 \over \sqrt{a^2+h^2}}\right) = {2GMm\over a^2}\cdot \left(1-{1 \over \sqrt{(\frac{a}{h})^2+1}}\right)$$

Using $x = \dfrac{a}{h} < 1$, we have to expand $\dfrac{1}{\sqrt{1 + x^2}} = 1 - \dfrac{x^2}{2}+O(x^4)$ to two terms using the Binomial series.

So we have:

$$F \sim {2GMm\over a^2}\cdot \left(1-\left(1-\frac{1}{2}\left(\frac{a}{h}\right)^2\right)\right) = {GMm\over h^2}$$