How Does Gauss' Law Apply to Gravitational Fields?

[arl146]

Homework Statement

The gravitational field g due to a point mass M may be obtained by analogy with the electric field by writing an expression for the gravitational force on a test mass, and dividing by the magnitude of the test mass, m. Show that Gauss' law for the gravitational field reads:

\Phi = \oint g\bullet dA = -4*pi*G*M

Use this result to calculate the gravitational acceleration g at a distance of R/2 from the center of a planet of radius R = 8.05 × 10^6 m and M = 8.45 × 10^24 kg.

Homework Equations

above equation

The Attempt at a Solution

i can't get the answer right for this .. here's what i did

\Phi = \oint g\bullet dA = -4*pi*G*M
g\oint dA = -4*pi*GM
g[4*pi*r^2] = -4*pi*GM
g[4*pi*(R/2)^2] = -4*pi*GM
g*pi*R^2 = -4*pi*GM
g = (-4GM)/R^2

and since r=R/2 the mass is halved also. therefore g = (-2*G*M)/R^2

i plugged in the values for G, M, and R .. and got -17.40267737 m/s^2 but its not right

 

[arl146]

anybody can give any hints of what I am doing wrong?

 

[Doc Al]

and since r=R/2 the mass is halved also.

Are you sure about that?

(What percentage of the sphere's volume--and thus mass, assuming uniform density--is located at r < R/2?)

 

[arl146]

ummm .. is the mass 1/8 of M? since V= (4/3)*pi*r^3
and since R=r/2 ... that makes it V = (4/3)*pi*(R^3/8)
meaning the volume is 1/8 of the total. and since D = M/V ---> M=DV so the mass also is 1/8 of the original?

 

[SammyS]

ummm .. is the mass 1/8 of M? since V= (4/3)*pi*r^3
and since R=r/2 ... that makes it V = (4/3)*pi*(R^3/8)
meaning the volume is 1/8 of the total. and since D = M/V ---> M=DV so the mass also is 1/8 of the original?

Assuming that the density of the planet is uniform, yes, that's correct.

 

[arl146]

is g supposed to be negative? also i got 4.3506693 m/s^2 is that right can someone check for me?

 

[Doc Al]

is g supposed to be negative? also i got 4.3506693 m/s^2 is that right can someone check for me?

That looks good. g is negative just means that the field points toward the center.

 

[arl146]

ok i got it that makes sense. thanks!