For short distances $r\ll \lambda$ calculate the force between $m$ and $m'$.
- $F=-\frac{Gmm'}{r^2}$
- $F=-\frac{Gmm'}{r^2}(1-a)$
- $F=-\frac{Gmm'}{r^2}(1+a)$
- $F=-\frac{Gmm'}{\lambda r}$
\begin{eqnarray*}
-V^\prime(r) & = & -\frac{Gmm'}{r^2}(1-ae^{-r/\lambda})-\frac{Gmm'}{r}(ae^{-r/\lambda}) \\
& = & -\frac{Gmm'}{r^2}\left(1-ae^{-r/\lambda}\left(1+\frac{r}{\lambda}\right)\right)
\end{eqnarray*}\]
When $r\ll \lambda$, we have \[F(r)\approx -\frac{Gmm'}{r^2}(1-a).\]
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r/lambda = 0 then we get 1-a(e^0) = 1-a
we can do this because r is MUCHMUCH smaller than lambda
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