Question

If a second planet were of the same radius R and made of the same material but had a hollow center of radius 0.50 R , what would be the acceleration of gravity at its surface

Answer 1

Answer:

$g_2=\dfrac{7}{8}g$

Explanation:

G = Gravitational constant

M = Mass of planet

R = Radius of planet

Acceleration due to gravity on first planet

$g=\dfrac{GM}{R}$

Assuming that the planets have the same mass density $\rho$

Density of first planet

$\rho=\dfrac{M}{V}\\\Rightarrow \rho=\dfrac{M}{\dfrac{4}{3}\pi R^3}\\\Rightarrow M=\rho \dfrac{4}{3}\pi R^3$

Density of second planet

$\rho=\dfrac{M_2}{V_2}=\dfrac{M_2}{\dfrac{4}{3}\pi R^3-\dfrac{4}{3}\pi (0.5R)^3}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{2^3})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{8})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(\dfrac{7}{8})}\\\Rightarrow M_2=\rho\dfrac{4}{3}\pi R^3(\dfrac{7}{8})\\\Rightarrow M_2=M\dfrac{7}{8}$

Acceleration due to gravity on the second planet

$g_2=\dfrac{GM_2}{R}\\\Rightarrow g_2=\dfrac{GM\dfrac{7}{8}}{R}\\\Rightarrow g_2=\dfrac{7}{8}g$

The acceleration due to gravity of the planet would be $\dfrac{7}{8}$ times the acceleration due to gravity on the first planet.