How Does Gauss' Law Apply to Gravitational Fields?

In summary, the gravitational field due to a point mass 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. Gauss' law for the gravitational field reads:
  • #1
arl146
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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:

[itex]\Phi[/itex] = [itex]\oint g\bullet dA[/itex] = -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

[itex]\Phi[/itex] = [itex]\oint g\bullet dA[/itex] = -4*pi*G*M
g[itex]\oint dA[/itex] = -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
 
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  • #2
anybody can give any hints of what I am doing wrong?
 
  • #3
arl146 said:
and since r=R/2 the mass is halved also.
Are you sure about that? :wink:

(What percentage of the sphere's volume--and thus mass, assuming uniform density--is located at r < R/2?)
 
  • #4
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?
 
  • #5
arl146 said:
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.
 
  • #6
is g supposed to be negative? also i got 4.3506693 m/s^2 is that right can someone check for me?
 
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  • #7
arl146 said:
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.
 
  • #8
ok i got it that makes sense. thanks!
 

What is Gauss' Law for Gravitation?

Gauss' Law for Gravitation is a mathematical law that describes the relationship between the gravitational force and the distribution of mass within a closed surface. It states that the flux of the gravitational field through any closed surface is proportional to the enclosed mass.

How is Gauss' Law for Gravitation used?

Gauss' Law for Gravitation is used to calculate the gravitational field at a point due to a distribution of mass. It allows us to simplify complex gravitational systems by considering the overall mass distribution rather than individual masses.

What is the significance of Gauss' Law for Gravitation?

Gauss' Law for Gravitation is significant because it is a fundamental law of physics that helps us understand the behavior of the universe. It is also used in many applications such as predicting the orbits of planets and spacecraft.

Is Gauss' Law for Gravitation the same as Coulomb's Law?

No, Gauss' Law for Gravitation and Coulomb's Law are two different laws that describe different types of forces. Coulomb's Law describes the electrostatic force between two charged particles, while Gauss' Law for Gravitation describes the gravitational force between two masses.

Can Gauss' Law for Gravitation be applied to all situations?

Gauss' Law for Gravitation is a general law that applies to all situations involving gravitational forces. However, it is most useful in situations where the mass distribution is symmetric and the gravitational field is constant.

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