Question

One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R, and mass 4M. If M = 6.24 × 10^24 kg and R = 4.11 × 10^6 m, what is the gravitational acceleration of a particle at points (a) R and (b) 3R from the center of the planet?

Answer 1

(a) $24.6 m/s^2$

At a distance r=R from the centre of the planet, there is no effect due to the outer shell: so, the gravitational field strength at r=R is only determined by the gravity produced by the core of the planet.

So, the strength of the gravitational field is given by

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

where

G is the gravitational constant

M = 6.24 × 10^24 kg is the mass of the core of the planet

R = 4.11 × 10^6 m is the radius of the core

Substituting into the equation, we find

$g= \frac{(6.67\cdot 10^{-11})(6.24\cdot 10^{24} kg)}{(4.11\cdot 10^6 m)^2}=24.6 m/s^2$

(b) $13.7 m/s^2$

at distance r=3R from the centre, the particle feels the effect of gravity due to both the core of the planet and the outer shell between R and 2R.

So, we have to consider the total mass that exerts the gravitational attraction at r=3R, which is the sum of the mass of the core (M) and the mass of the shell (4M):

M' = M + 4M = 5M

Therefore, the gravitational acceleration at r=3R will be

$g'= \frac{G(5M)}{(3R)^2}=\frac{5}{9}\frac{GM}{R^2} = \frac{5}{9}g$

And susbstituting

g = 24.6 m/s^2

found in the previous part, we find

$g' = \frac{5}{9} (24.6 m/s^2)=13.7 m/s^2$