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

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Homework Help Overview

The discussion revolves around finding the acceleration due to gravity inside a spherical planet of uniform density. The original poster presents a problem involving the gravitational acceleration at a distance from the planet's center, questioning the use of variables and the application of relevant equations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive an expression for gravitational acceleration inside the planet and expresses confusion regarding the use of the radius in the equations. Some participants question the application of Newton's Shell Theorem and the clarity of variable usage in the original post.

Discussion Status

The discussion is ongoing, with participants providing hints and questioning the assumptions made by the original poster. There is an acknowledgment of the need for clearer variable definitions and the relevance of established theorems in understanding the problem.

Contextual Notes

There is a noted ambiguity in the use of the variable names, which may lead to confusion in understanding the problem setup. The original poster's reference to previously seen questions indicates a familiarity with the topic but also highlights uncertainty in grasping the underlying concepts.

Johnnie123
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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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I don't know the math so I can't tell but - are you taking into account Newton's Shell Theorem?
 
Johnnie123 said:
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.
 
PeroK said:
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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