# Derive gravitational potential energy for this system [closed]

This is on a study guide for my Physics 221 final. I feel like I almost got it but I am off by a sign error. Here is the question:

Here is what I got so far:

Known:

$$F_g = \frac{GMm}{r^2}$$ $$U_g = -\int F\space dr$$

My work:

$$U_g = -\int_\sqrt{x^2+a^2}^\infty \frac{GMm}{r^2}\space dr$$ $$=\frac{GMm}{r} |_\sqrt{x^2 + a^2}^\infty$$ $$=\frac{GMm}{\infty} - \frac{GMm}{\sqrt{x^2 + a^2}}$$ $$=-\frac{GMm}{\sqrt{x^2 + a^2}}$$

My answer is super close but it is just off by a sign. If I flip the bounds of my integral my answer works, but if that is what I do I don't understand why. My bounds make sense in my head. Where is my mistake?

• The sign in the question seems wrong, the energy should drop as the two masses come closer together.
– Ian
Commented Dec 7, 2019 at 0:56

As Cookie17 points out, since the mass source is a continuous distribution, then we must divide the ring into small pieces of mass $$dM$$, so the total gravitational potential energy of the system is given by \begin{align} U_g = -Gm \int_{M} \frac{1}{|\vec{x}-\vec{r}|} dM(\vec{r}), \end{align} where $$\vec{x}$$ is the position of the point-mass $$m$$ and $$\vec{r}$$ points towards each mass element of the ring. This way, $$|\vec{x}-\vec{r}|=\sqrt{x^2+a^2}$$ is the distance from the mass $$m$$ to each mass element. Since $$x$$ and $$a$$ are constant quantities, then we only integrate the mass element of the ring, giving us the total mass $$M$$. This is, \begin{align} U_g&=-\frac{Gm}{\sqrt{x^2+a^2}}\int_MdM(r), \\ &=-\frac{GmM}{\sqrt{x^2+a^2}}. \end{align} Note that the result has a minus sign. A nice way to check this is applying $$\vec{F}_g=-\nabla U_g$$ to the latter result considering a variable distance $$x$$, so you get a gravitational force exerted upon the point-mass pointing towards the ring. This is what physically happens since this force is always attractive.

Note: In this case the symmetry of the problem implies that the distance is constant. But, in general, one must integrate the expression considering a mass-density $$\rho$$ such that $$dM(\vec{r})=\rho(\vec{r})dV$$, where $$dV$$ is a volume-element in a 3D mass distribution.

A serious error that you have made is to treat the gravitational force between the ring and the test mass $$\vec F_{\rm grav}$$ as a scalar.

What you need, when you do your integration to find the work done by the gravitational field, is to use the component of the force in the $$\hat x$$ direction, $$-\dfrac{GMm}{(a^2+x^2)}\,\cos \theta$$.
The total mass of the ring, $$M$$, is used because by symmetry each element of the ring makes an equal contribution to the component of the force in the $$\hat x$$ direction.

To find the gravitational potential energy of the ring and test mass system you now have to evaluate the integral, $$-\int _\infty ^x \left [-\dfrac{GMm}{(a^2+x^2)}\,\cos \theta \,\hat x \right]\,\cdot dx \,\hat x$$
Note the limits of integration with the test mass moving from infinity to position $$x$$.

The potential energy will be negative.
This is what you might expect because as the test mass approaches the ring the system has less potential energy than the zero amount it had when the masses were infinitely far apart.

• I like the idea of Symmetry that makes sense to me, but if you evaluate that integral you get a very complicated result that does not resemble the "correct" answer given by the question. Am i missing something? Commented Dec 8, 2019 at 19:10
• @LukeKelly $\cos \theta = \frac{x}{\sqrt{a^2+x^2}}$ which when combined with the other term and integrated produces $- \frac{x}{\sqrt{a^2+x^2}}$. Commented Dec 8, 2019 at 20:33

You are treating the ring like it is a point particle when it is not. You will need to split the ring into many dM's and find the potential energy from each dM and integrate. Each dM can be treated like a point particle and it should start like this

$$dU_g = -\frac{Gm}{r}\space dM$$

You can use mass density to find dM in terms of ds, a small portion of arc length. You can then integrate both sides, setting limits of integration from 0 to the circumference of the ring.

• Would this require a double integral? Commented Dec 7, 2019 at 1:23
• No just a single integral Commented Dec 7, 2019 at 1:35