Explanation of 17.5 #48

In this problem, you are asked to compute the gravitational potential from uniform density spherical shell of total mass \(m\) and radius \(R\). In general, the gravitational potential of a massive surface \(S\) is given by
\[
V(a, b, c) = – G \iint_S \frac{\rho, dS}{\sqrt{(x – a)^2 + (y – b)^2 + (z – c)^2}}
\]
where \(\rho\) is the density (mass per area) of the surface. In this case, \(\rho = m / (4 \pi R^2)\).

The first part of the question asks you to reduce the problem to the case where \((a, b, c) = (0, 0, r)\). We can do this by spherical symmetry: given any \((a,b,c)\) we can rotate the entire picture so that the point \((a, b, c)\) lies on the positive \(z\) axis. That is, \((a, b, c) = (0, 0, r)\) with \(r = \sqrt{x^2 + y^2 + z^2}\). Applying this rotation doesn’t change the gravitational potential because the rotated sphere is indistinguishable from the original sphere.

The second part of the question asks you to set up the integral for \(V(0, 0, r)\) in spherical coordinates. In spherical coordinates (with radius \(R\)) we have
\[
x = R \cos \theta \sin \varphi, \quad y = R \sin \theta \sin \varphi, \quad z = R \cos \varphi.
\]
You should verify that the normal vector for this parametrization satisfies \(\|n\| = R^2 \sin \varphi\). Notice that the denominator of the integral defining the gravitational potential is just the distance from \((a, b, c)\) to \((x, y, z)\) (where the latter is a point on the surface). In the case where \((a, b, c) = (0, 0, r)\) and \(S\) is the sphere of radius \(R\) centered at the origin, we can compute this distance using the law of cosines as indicated in the following figure:
Law Of Cosines
In the figure, \(d\) is the distance from the point \(P = (0,0,r)\) to a point \(Q = (x, y, z)\) lying on the sphere — that is, \(d\) is the denominator of the integral we’re trying to compute. By the law of cosines,
\[
d^2 = r^2 + R^2 – 2 R r \cos \varphi.
\]
Therefore, the potential is given by
\[
V(0, 0, r) = \frac{- G m}{4 \pi} \int_0^{\pi} \int_{0}^{2 \pi} \frac{\sin \varphi , d\theta, d\varphi}{\sqrt{R^2 + r^2 – 2 R r \cos \varphi}}.
\]
The book suggests using the substitution \(u = R^2 + r^2 – 2 R r \cos \varphi\) to compute this integral.

Will Rosenbaum

Tel Aviv

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