Kevin S. Huang

2014: Problem 17

Topic: Gravity
Concepts: Gauss’s law for gravity

Solution:

By the shell theorem, the gravitational field outside a uniform sphere is the same as that of a point at the center of the sphere with the same mass. Thus,

g=\frac{GM_{\text{enclosed}}}{R^2} \propto \frac{G(\rho R^3)}{R^2} \propto \rho R

Since the radius is fixed, we have $g \propto \rho$ so it remains to find how the density changes. When the cloud expands to radius $2R_0$ , the same mass is spread over a larger volume. We have

M=\rho_0\left(\frac{4}{3}\pi R_0^3\right)=\rho’\left[\frac{4}{3}\pi(2R_0)^3\right]

\rho_0=8\rho’

Since the new density $\rho’=\rho_0/8$ ,

g’=g_0/8

so the answer is C.

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Kevin S. Huang

2014: Problem 17

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2014: Problem 17

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