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
$$F = -\frac{Gmm'}{r^2}(1-a)$$

Powered by Hirophysics.com