The matter that makes up a planet is distributed uniformly so that the planet has a fixed, uniform density. How does the magnitude of the acceleration due to gravity g at the planet surface depend on the planet radius R ? (Hint: how does the total mass scale with radius?) g ∝ 1 / R g ∝ R g ∝ √ R g ∝ R 2

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alexdott3023 alexdott3023

03-12-2019
Physics

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The matter that makes up a planet is distributed uniformly so that the planet has a fixed, uniform density. How does the magnitude of the acceleration due to gravity g at the planet surface depend on the planet radius R ? (Hint: how does the total mass scale with radius?) g ∝ 1 / R g ∝ R g ∝ √ R g ∝ R 2

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lucianoangelini92 lucianoangelini92 · Answer 1 · 2019-12-04T16:11:26+01:00

Answer:

the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)

Explanation:

according to newton's law of universal gravitation ( we will neglect relativistic effects)

F= G*m*M/d² , G= constant , M= planet mass , m= mass of an object , d=distance between the object and the centre of mass of the planet

if we assume that the planet has a spherical shape, the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³ ,

since M= ρ* V = ρ* 4/3πR³ , ρ= density

F = G*m*M/R² = G*m*ρ* 4/3πR³/R²= G*ρ* 4/3πR

from Newton's second law

F= m*g = G*ρ*m* 4/3πR

thus

g = G*ρ* 4/3π*R = (4/3π*G*ρ)*R

g ∝ R