With what force $M$ distributed with uniform density over the circle $x^2 + y^2 = a^2$, $z = 0$ act on $m$ located at the point $A(0,0,b)$.

Question

With what force will a mass $M$ distributed with uniform density over the circle $x^2 + y^2 = a^2$, $z = 0$ act on a mass $m$ located at the point $A(0,0,b)$.

I want to setup the problem as a line integral where $C:x^2+y^2 = a^2$. Parameterizing via $x = a \cos(\theta), y = a\sin(\theta)$

$$\int_C Fds = \int_{0}^{2\pi}\frac{Gm}{r^2}dM$$

The $r$ will always be $\sqrt{a^2+b^2}$. For a circle $dM = \frac{2\pi a}{M}ds = \frac{2\pi a}{M} a d\theta$. Thus resulting in:

$$\int_{0}^{2\pi} \frac{Gm}{a^2 + b^2}2\pi a^2 d\theta = \frac{4\pi^2a^2Gm}{a^2 + b^2}$$

Is this answer correct?

Update - Ok apparently the answer is $$ \frac{GMmb}{(a^2 + b^2)^{3/2}}$$

Answer 1

The force over mass $m$ at $(0, 0, b)$ by the mass $M$ uniformly distributed over the circle $r = a, z= 0$ can be written as,

$ \displaystyle \int_{0}^{2\pi}\frac{G \ m \cos\phi}{r^2} \ dM $

You missed $\cos\phi$ which is to consider force only along z-axis as the force along xy-plane will cancel out over the circle.

where $\cos\phi = \cfrac{b}{\sqrt{a^2+b^2}}$

$dM = \cfrac{M}{2\pi a} \ ds = \cfrac{M}{2\pi} \ d\theta$

So the integral is,

$ \cfrac{G \ m \ b}{(a^2+b^2)^{3/2}}\displaystyle \int_{0}^{2\pi} \cfrac{M}{2\pi} \ d\theta $

$ = \displaystyle \cfrac{G \ M \ m \ b}{(a^2+b^2)^{3/2}}$

Answer 2

Integrating a vector needs to be done by components.

Your first integral has the wrong units. It would work out to an energy.

I would have set up the first integral as $\overrightarrow F = \int d\overrightarrow F$. which leads to three component equations, only one of which is relevant: $F_z = \int dF_z$ The vertex half angle $\phi$ satisfies $\tan \phi = b/a$ and you can use that to express the $z$-component of the force.

Answer 3

Now that there are other answers with the correct answer, I'd like to post a slightly more explicit calculation so you can see how these sorts of calculations work in general.

We have $$ d\vec{F} (\vec{r}) = \frac{Gm\,dM(\vec{r}')}{\|\vec{r} - \vec{r}'\|^3} (\vec{r} - \vec{r}') $$ where $\vec{r}$ is the point at which we want to know the force and $\vec{r}'$ is the location of the mass $dM$, indicated above with the notation $dM(\vec{r}')$. We integrate $$ \vec{F} (\vec{r}) = \int \frac{Gm\,dM(\vec{r}')}{\|\vec{r} - \vec{r}'\|^3} (\vec{r} - \vec{r}') $$ Now, $$ \vec{r} - \vec{r}' = (0, 0, b) - (a \cos \theta, a \sin \theta , 0) = (a\cos\theta , a\sin\theta , b) $$ so $$ \| \vec{r} - \vec{r} ' \|^3 = (a^2 + b^2)^{\frac{3}{2}} $$ and: $$ dM(\vec{r}') = \frac{M}{2\pi a} ds = \frac{M}{2\pi} a d\theta = \frac{M}{2\pi} d\theta $$ We may now calculate: \begin{align} \vec{F} (\vec{r}) &= \int_0^{2\pi} \frac{Gm \left(\frac{M}{2\pi} d\theta\right)}{(a^2 + b^2)^{\frac{3}{2}}} (a \cos\theta , a \sin\theta, b) \\&= \frac{ G m M }{2\pi (a^2 + b^2)^{\frac{3}{2}}}\int_0^{2\pi} (a \cos\theta , a \sin\theta, b) d\theta \\&= \frac{ G m M }{2\pi (a^2 + b^2)^{\frac{3}{2}}}(0 , 0, 2\pi b) \\&= \left( 0 , 0 ,\frac{G m M b}{(a^2 + b^2)^{\frac{3}{2}}}\right) \end{align} Finally, the answer to your question is $$ \vec{F}(0, 0, b) = \left( 0 , 0 ,\frac{G m M b}{(a^2 + b^2)^{\frac{3}{2}}}\right) $$