# Simple gravitational force problem (Gregory Classical Mechanics)

1. Mar 3, 2013

### mathmonkey

1. The problem statement, all variables and given/known data

This is question 3.7 from Gregory's Classical Mechanics textbook.

A symmetric sphere of radius a and mass M has its center a distance b from an infinite plane containing a uniform distribution of mass $\sigma$ per unit area. Find the gravitational force exerted on the sphere

2. Relevant equations

3. The attempt at a solution

From what I understand, a solid sphere can be represented as if it were a single particle of mass M concentrated at its center of mass, call this point $S$. So, the way I approached the problem was summing (integrating) up the forces exerted on this point by each infinitesimally small area $dxdy$ on the plane.

The mass of each infinitesimal on the plane is $m = \sigma dA$. I also let $\theta$ represent the angle between SB (where B is the straight line distance from the point S) and the line drawn from S to the infinitesimal. Then, the equation I got was:

$F = MG \int _A \sigma \cos (\theta) / R^2 dA$
$F = MG\sigma \int _A R\cos (\theta) / R^3 dA$
$F= MG\sigma \int _A b/R^3 dA$
$F = MGb\sigma \int _{-\infty}^\infty \int _{-\infty}^\infty 1/(x^2 + y^2 + b^2)^{3/2} dxdy$

If anyone could let me know if I have set this up correctly (I get the feeling I have not), and how to approach this problem, I'd be really grateful. Thanks!

2. Mar 3, 2013

### voko

The reduction of the problem to that of a point mass is correct.

Then, as far as I can see, you integrate the parallel component of the force, ignoring the perpendicular component. I think you owe us an explanation here.

3. Mar 3, 2013

### mathmonkey

So from what I understand, if we drop a straight line from the center of the sphere S to where it meets the plane, say that point is O, we'll have a line SO perpendicular to the plane of length $b$. Then if we drop another straight line from S to a particular point on the plane, say X, then SOX becomes a right triangle where the angle created by the lines SO and SX is $\theta$. Also:

$|SO| = b$
$|OX| = R \sin(\theta)$
$|SX| = R$

Then, I supposed by symmetry, that all the component forces by each point $X$ in the plane will sum up to a single perpendicular force pointing in the direction of $SO$. So, if each point $X$ attracts S in its direction with a force of magnitude

$F' = m(\sigma dA)G/R^2$

Then its force in the direction of SO ought to be $F = F'\cos (\theta)$. That's how I derived my integral in the form of

$F = mG \int _A \sigma \cos (\theta)/R^2 dA$

4. Mar 3, 2013

### voko

There is no error in your derivation. It is entirely correct. You just need to finish integration.

5. Mar 3, 2013

### mathmonkey

Hmm, then maybe I am having trouble with simplifying the integral. What is the most efficient way to do it? What I thought to do is this:

$F = MG\sigma \int _A \cos (\theta) / R^2 dA$
$F = MG\sigma \int _A R\cos (\theta) /R^3 dA$
$F = MG\sigma \int _A b / R^3 dA$

After this, I assumed by the pythagorean theorem that $R = \sqrt{ x^2 + y^2 + b^2 }$, but then I end up with what looks like a nasty integral:

$F = MG\sigma b \int _{-\infty}^{\infty} \int _{-\infty}^{\infty} 1/ (\sqrt{x^2 + y^2 + b^2})^3 dxdy$

Is there another way to approach this, or am I simplifying the integral wrongly? Thanks!

6. Mar 3, 2013

### mathmonkey

Ah never mind, I've figured it out. The better way is to use polar coordinates for the area of the plane. The integral simplifies to

$F = MGb\sigma \int _{0}^{2\pi} \int _{0}^{\infty} r/(r^2+b^2)^{3/2}drd\phi$

Thanks for the help.