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How to find the acceleration due to gravity inside a planet?

  • #1
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Consider a spherical planet of uniform density ρ. The distance from the planet's center to its surface (i.e., the planet's radius) is R. An object is located a distance R from the center of the planet, where R < Rp. (The object is located inside of the planet.)

1) Find an expression for the magnitude of the acceleration due to gravity, g(R), inside the planet.

2) Rewrite your result for g(R) in terms of gp, the gravitational acceleration at the surface of the planet, times a function of R.

Relevant equations:

$$g = \frac{GM}{R^2}$$


So, I have seen this question on the forum before, but I am still unsure about the concept.

For 1, I found $$g(R) = \frac{Gp(4/3)*π*R^3}{R^2}$$, which is the correct answer. However, I do not understand why R^3 in the numerator is R^3 and not (R_p)^3. In the equation $$g = \frac{GM}{R^2}$$ M is the entire mass of the planet or spherical body. Wouldn't that require $$M=p*(4/3)*π*R^3$$
 
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Answers and Replies

  • #2
DaveC426913
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I don't know the math so I cant tell but - are you taking into account Newton's Shell Theorem?
 
  • #3
PeroK
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Consider a spherical planet of uniform density ρ. The distance from the planet's center to its surface (i.e., the planet's radius) is R. An object is located a distance R from the center of the planet, where R < Rp. (The object is located inside of the planet.)

1) Find an expression for the magnitude of the acceleration due to gravity, g(R), inside the planet.

2) Rewrite your result for g(R) in terms of gp, the gravitational acceleration at the surface of the planet, times a function of R.

Relevant equations:

$$g = \frac{GM}{R^2}$$


So, I have seen this question on the forum before, but I am still unsure about the concept.

For 1, I found $$g(R) = \frac{Gp(4/3)*π*R^3}{R^2}$$, which is the correct answer. However, I do not understand why R^3 in the numerator is R^3 and not (R_p)^3. In the equation $$g = \frac{GM}{R^2}$$ M is the entire mass of the planet or spherical body. Wouldn't that require $$M=p*(4/3)*π*R^3$$
Your post is difficult to understand because you seem to use the same letters for different quantities, starting with ##R## being the radius of the planet and ##R## also being the variable distance from the centre.

You could use ##r## for the variable radius and ##m## for the mass enclosed by a shell of radius ##r##.

In answer to your question, when you are inside the planet, is the entire mass of the planet pulling the object towards the centre?

The post above, pointing you at the shell theorem is a good hint.
 
  • #4
DaveC426913
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Your post is difficult to understand because you seem to use the same letters for different quantities, starting with ##R## being the radius of the planet and ##R## also being the variable distance from the centre.
Ah. OK. I thought maybe what he was demoing was just beyond me.:smile:
 

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