Question:

Newtonian theory of gravity can be modified at short range. Such potential energy between two objects is given as

\[  U(r) = -\frac{Gmm’}{r}(1-ae^{-\frac{r}{\lambda}})  \]

Find 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.

\[  F = -\frac{dU}{dr}  \]

\[   = -\frac{Gmm’}{r^2}(1-ae^{-\frac{r}{\lambda}})+\frac{Gmm’}{r}\frac{a}{\lambda}(1-ae^{-\frac{r}{\lambda}}) \]

\[   = -\frac{Gmm’}{r^2}\left(1-ae^{-\frac{r}{\lambda}}\left(1+\frac{r}{\lambda}\right)\right)  \]

Since \(r \ll \lambda\), \(\frac{r}{\lambda}\approx 0\). Therefore, the force for short ranges is

\[  F = -\frac{Gmm’}{r^2}(1-a)  \]