(a) [tex]24.6 m/s^2[/tex]
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
[tex]g= \frac{GM}{R^2}[/tex]
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
[tex]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[/tex]
(b) [tex]13.7 m/s^2[/tex]
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
[tex]g'= \frac{G(5M)}{(3R)^2}=\frac{5}{9}\frac{GM}{R^2} = \frac{5}{9}g[/tex]
And susbstituting
g = 24.6 m/s^2
found in the previous part, we find
[tex]g' = \frac{5}{9} (24.6 m/s^2)=13.7 m/s^2[/tex]
