Physics 9
Spring 2012
TA:
NAME:
Quiz 3 - Solutions
Make sure your name is on your quiz, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
Because the formulas for Coulomb’s law and Newton’s law of gravity have the same
inverse-square law dependence on distance, a formula analogous to the formula for Gauss’s
law can be found for gravity. The gravitational field ~g at a location is the force per unit
mass on a test mass m0 placed at that location. (Then, for a point mass m at the origin,
the gravitational field g at some position ~r is ~g = − (GN m/r2 ) r̂.)
1. Compute the flux of the gravitational field through a spherical surface of radius r, and
verify that the gravitational analog of Gauss’s law is
I
~ = −4πGN mencl .
Φg = ~g · dA
2. Determine the gravitational field, ~g , a distance r from the center of the Earth (where
r ≤ RE , where RE is the radius of the Earth), assuming that the Earth’s density is
uniform. Express your answer in terms of GN , the mass of the Earth, ME , and RE .
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Solution
1. The gravitational field from a spherical mass distribution is again spherical. Hence,
the field will be constant on a spherical Gaussian surface. Thus, taking our Gaussian
surface as a sphere of radius r, we have
I
~ = g 4πr2 .
~g · dA
If this Gaussian surface encloses a point mass m, then mencl = m, and so
GN m
g 4πr2 = −4πGN m ⇒ g = − 2 ,
r
which is the correct field.
2. We can determine the gravitational field in this case in exactly the same way as we
would determine the electric field inside a uniformlyHcharged sphere. Taking our Gaus~ = g (4πr2 ). Then, in terms
sian surface as a sphere of radius r, again, we have ~g · dA
3
of the density, ρ = ME /VE , where ME is the mass of the Earth, and VE = 4πRE
/3 is
4πr3
the volume, the enclosed mass can be written mencl = 3 ρ. Thus,
4πGN ρ
4πr3
ρ⇒g=−
r,
g 4πr2 = −4πGN
3
3
or in terms of the mass, ME ,
GN ME
~g = −
rr̂.
3
RE
1