Gravity: Force and potential energy

Question:
Newtonian theory of gravity can be modified at short range. The potential energy between two objects is given as
\[
U(r) = -\frac{Gmm'}{r}(1-ae^{-\frac{r}{\lambda}})
\]
What is the force between $m$ and $m'$ for short distances ($r \ll \lambda$)?

Answer:
The force is conservative and calculated by the derivative of the potential energy with respect to the distance.
\begin{eqnarray}
F &=& -\frac{dU}{dr}  \\
 &=& -\frac{Gmm'}{r^2}(1-ae^{-\frac{r}{\lambda}})+\frac{Gmm'}{r}\frac{a}{\lambda}(1-ae^{-\frac{r}{\lambda}}) \\
 &=& -\frac{Gmm'}{r^2}\left(1-ae^{-\frac{r}{\lambda}}\left(1+\frac{r}{\lambda}\right)\right)
\end{eqnarray}
Since $r \ll \lambda$, $\frac{r}{\lambda}\approx 0$. Therefore, the force for short ranges is
\[
F = -\frac{Gmm'}{r^2}(1-a)
\]

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