# Force derived from Yukawa potential

This is with regards to problem 3.19 from Goldstein's Classical Mechanics,

A particle moves in a force field described by the Yukowa potential $$V(r) = -\frac{k}{r} e^{-\frac{r}{a}},$$ where $$k$$ and $$a$$ are positive.

where I bolded the assumptions as this is the only information I can imagine helps me resolve this.

A solution due to Professor Laura Reina at Florida State Uni, as well as a solution due to Slader.com

both use the following expression for the force felt by a particle in the given Yukawa potential:

$$F(r) = -\frac{k}{r^2} e^{-\frac{r}{a}}$$

I am struggling to wrap my head around this. This is clearly not the result of

$$-\frac{\partial V(r)}{\partial r}$$

which evaluates to

$$-\frac{k}{r^2} e^{-\frac{r}{a}} - \frac{k}{ar} e^{-\frac{r}{a}}$$

Can anyone help me understand why the second term $$-\frac{k}{ar} e^{-\frac{r}{a}}$$ can be excluded here? I tried plotting some various example of this, varying k and a which are allowed to be any positive numbers, but I've no insight.

There was a question regarding this same topic which was not answered Deriving potential from central force

• I looked at Reina’s solution. Her $F(r)$ is wrong but she isn’t using it. The force in the 3rd equation on page 5 is correct. – G. Smith Jul 25 '20 at 22:35
• Your $V(r)$ is a repulsive potential and doesn’t have orbits. Doesn’t Goldstein have a negative sign? Reina is using $V(r)=-\frac{k}{r}e^{-r/a}$. – G. Smith Jul 25 '20 at 22:40
• And be careful: her $V’(r)$ doesn’t mean $dV/dr$! Very confusing! – G. Smith Jul 25 '20 at 22:46
• Don’t forget that $F_r=-dV/dr$ has a negative sign. – G. Smith Jul 25 '20 at 22:50
• Thank you very much for the clarification! The pointer to the correction expression on page 5 helps! :D and indeed I'm missing a minus sign! – Lopey Tall Jul 25 '20 at 23:29

I think the solutions you have posted are just incorrect. The exact expression should include the $$-\frac{k}{ar}e^{-ar}$$ term. You could approximately ignore this term in the limit $$r\ll a$$, but in this limit one should also Taylor expand the exponential to be consistent.