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Homework Help: Gravitational Acceleration inside a Planet

  1. Nov 27, 2007 #1
    The problem:

    Consider a spherical planet of uniform density [tex]\rho[/tex]. The distance from the planet's center to its surface (i.e., the planet's radius) is [tex]R_{p}[/tex]. An object is located a distance [tex]R[/tex] from the center of the planet, where [tex]R\precR_{p}[/tex] . (The object is located inside of the planet.)

    Part A

    Find an expression for the magnitude of the acceleration due to gravity, [tex]g(R)[/tex] , inside the planet.

    Express the acceleration due to gravity in terms of [tex]\rho[/tex], [tex]R[/tex], [tex]\pi[/tex], and [tex]G[/tex], the universal gravitational constant.

    Part B

    Rewrite your result for [tex]g(R)[/tex] in terms of [tex]g_{p}[/tex], the gravitational acceleration at the surface of the planet, times a function of R.

    Express your answer in terms of [tex]g_{p}[/tex], [tex]R[/tex], and [tex]R_{p}[/tex].

    My attempt at a solution:

    I determined the answer to Part A to be [tex]g(R)=(4/3)G\rho \pi R[/tex]. However, I am uncertain how to find the answer to Part B. I barely even understand what they are asking me to do. I could really use some hints to point me in the right direction.

  2. jcsd
  3. Nov 27, 2007 #2

    Shooting Star

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    Homework Helper

    They want you to eliminate G, rho etc, and express the ans you got in terms of g at surface.

    You do know that M = (4/3)pi*Rp^3*rho. Also, you should know g at surface using law of gravitation. Use all these to eliminate the unwanted stuff.
  4. Nov 27, 2007 #3
    Ok, well I've tried to work this out, but I'm basically just guessing at everything--I'm that clueless. I don't even see how knowing M will help me. I don't know what to do.
  5. Nov 27, 2007 #4

    D H

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    What the question is asking you to do is to find some function [itex]f[/itex] such that

    [tex]g(R) = f(g_p,R_p,R)[/tex]

    In other words, somehow replace the [itex]G[/itex] and [itex]\rho[/itex] from the solution already at hand,

    [tex]g(R) = \frac 4 3 G \pho \pi R[/tex]

    with [itex]g_p[/itex] and [itex]R_p[/itex]. What is [itex]g_p[/itex]?
  6. Nov 28, 2007 #5

    Shooting Star

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    Put Rp in place of R in the formula you derived in our first post. Remember, g(Rp) is the g_p at the surface. So, you can write g(R) in terms of g_p and R.
  7. Nov 16, 2008 #6
    i'm doing the same question, got the first part right and i gotta admit, i still don't get it, i know it has something to do with substiting the value of g_p but and that that can be obtained by using the universal law of gravitation, but after that i am stumped.
  8. Nov 16, 2008 #7
    just worked it out, you gotta subsitute formulae and you should end up with R*g_p/R_p
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