# Gravitational force of a ring [closed]

Given a ring with mass $m_1$, radius $R$ and a point mass $m_2$ with distance $x$ regarding the center of the ring.

How can I calculate the gravitational force of this ring ? At first glance I'd say it is $F=\frac{m_1m_2}{s^2}G$ since every mass point of the ring is $s$ away from $m_2$, but I am pretty sure this is not correct.

My next idea is to consider a tiny mass piece $dm_1$ of the ring and then integrate over the whole ring, but here I struggle with the fact that the ring isn't considered 3-dimensional.

Due to symmetry $F_y$ (vertical force) has to be $0$.

Has anyone a hint? Also I would appreciate some general tips how to solve such kind of problems.

(Yes it is a homework, but as you can see, I want general tips and not the whole solution presented) • Mathematically this is exactly the same problem as you would have if this was a ring of charge. Do you know how to solve that problem? Knowing that by symmetry the final force needs to point along the x direction, the vertical components cancel - so find the horizontal component due a little bit of mass and integrate around the ring. Mar 26, 2015 at 17:08

Instead of integrating the force due to each mass element, which requires you to compute the component in the x-direction, you can calculate the gravitational potential, which is a scalar quantity. The force is then minus the gradient of the potential. The gravitational potential energy is:

$$V(x) = -\frac{m_1 m_2 G}{\sqrt{x^2 + R^2}}$$

The force in the x-direction is thus given as:

$$F(x) = -\frac{dV}{dx} = -G\frac{m_1 m_2 x}{\left(x^2 + R^2\right)^{3/2}}$$

• Why is $V(x) = - \frac{m_1m_2G}{\sqrt{x^2+R^2}}$ ? Mar 26, 2015 at 17:04
• The potential energy of two masses m1 and m2 at a distance r is -m1 m2 G/r. In case of the ring, all the mass elements of the ring are at the same distance sqrt(x^2 + R^2) from the point mass. Mar 26, 2015 at 17:36

Calculating the gravitational force on the axis of a ring is equivalent to calculating the gravitational force of a pair of opposing small portions of the ring in which the full mass $M$ of the ring is thought to be concentrated. The result is an axial force equal to

$$F = G M m\frac{cos(\theta)}{S^2}$$

Where $\theta$ is the half-angle between the two masses as seen from the position of the test mass $m$. Substituting $cos(\theta)=x/S$ it follows that

$$F= G M m \frac{x}{S^3}$$

• Ok, how do you get to the first equation ? Considering the tiny mass piece: $dF = \int Gm \frac{\cos(\phi)}{S^2}dM'$ from 0 to M ? Mar 26, 2015 at 17:20
• @Christian - just vector-sum the gravitational force of two diametrically opposite point masses of mass $\frac12 M$ each. Mar 26, 2015 at 17:22

$F=\frac{Gm_1m_2}{r^2}$ is wrong because you are essentially summing up the magnitude of the gravitational forces of each point on the ring without considering their direction. This is wrong since force is a vector.

To find the force, take 2 small elements, diametrically opposite on the ring each of length $dl$ and mass $dm$. Draw the direction of the force due the 2 elements on mass $m_2$. You will see that the vertical components cancel out. Take the sum of the horizontal components and integrate over the whole ring. The fact that the ring isn't 3 dimensional simplifies the problem as you don't have to worry about the element $dl$ having any thickness.

$$\vec{F} = - G m \int \frac{\varrho(\vec{r}\ ') \cdot (\vec{r} - \vec{r}\ ')}{|\vec{r} - \vec{r}\ '|^3} d^3r'$$ where $\varrho$ is the mass distribution. The Integral is taken over the whole volume (or equally: the support of $\varrho$).

For mass distributions, that have no 3D-volume, you have to take some Delta-terms, e.g. $$\varrho(x,y,z) = M\cdot \delta(x) \delta(y) \delta(z)$$ for a point mass at the origin.