00:02
In this exercise, we're asked to find the potential due to a uniform sphere.
00:10
So we're given s a hollow sphere of radius r with center at the origin, with a uniform mass distribution of a total mass, little m, the gravitational potential v due to the surface s at a point p with coordinates a, b, c is equal to negative and the gravitational constant big g times the surface integral over the surface s of the mass density row over the square root of x minus a squared plus y minus b squared plus z minus c squared.
01:12
Now as an aside, because our surface has surface area 4 pi r squared, the mass density row is equal to m over 4 pi r square.
01:29
In part a, we're asked to show that the potential depends only on the distance r from the point p to the center of the sphere.
01:50
The gravitational potential due to the surface s at a point p is given by, as we already pointed out, v of p equals negative g times the surface integral over for s of row over the square root of x minus a squared plus y minus b squared plus z minus c squared.
02:23
And of course sphere is symmetric.
02:34
By the symmetry of our sphere, the sum of the distances of point p, which is a, b, c, from point points q with coordinates x, y, z on our sphere s, is going to be equal for all points p located at some distance little r from the center of our sphere s.
04:16
So it follows at the integral over s of 1 over the length of q minus p depends only on little r.
04:50
Now, our mass density row is a constant.
05:01
Therefore, it follows that our integral for v, and therefore, v itself depends only on r.
05:32
This is what we wanted to show in part a.
05:38
Notice that this implies that we can simply compute v of p for a point p on the z axis.
06:02
So assume p is on the z axis at the point zero zero r, where little r is equal to the distance of p from the origin, squared of a squared plus b squared plus c squared, and little r of course is not equal to big r.
06:45
Next, in part b, we're asked to evaluate this surface integral, to show that it's equal to an iterated double integral using spherical coordinates.
07:06
Well, you have that v of 0, 0, 0, at point on the z axis.
07:18
This is ab equal 0 and c equal to r.
07:22
So this is going to be negative g times the double integral, or the surface integral over s of, or mass density row over, and then here this becomes the square root of x squared plus y squared plus z minus little r squared.
07:47
We can parameterize our sphere using the spherical coordinates.
07:54
So i'll call this, you already have g, so let's make this phi.
08:01
Capital phi of theta, lowercase, phi, is equal to the radius of our sphere, which is r, cosine, theta, sine little phi, r, sine theta, sine little phi, and r, cosine, little phi.
08:31
And, as usual, theta lies between 0 and 2 pi, and little phi.
08:43
And the length of the normal vector, n, in spherical coordinates, is the radius big r squared, times the sine of little phi.
08:56
Now let's express our function in the integrand in terms of our parameters.
09:01
So we have, first of all, that x squared plus y squared plus z minus r squared.
09:09
Well, this simplifies to r squared cosine squared theta, sine squared little phi plus r squared sine squared theta sine squared phi plus and then z minus little r is big r cosine phi minus little r squared now multiplying things out we get r squared and using triggered entity is r squared sine squared phi plus r squared cosine squared phi, it's big r squared, cosine squared, phi, minus two big r, little r, cosine, cosine, phi, plus little r squared...