Question

The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?

Answer 1

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

$g = \frac{Gm}{r^2}$

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

$g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2$ --------------------- equation (1)

FOR PLANET 2:

$g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}$

using equation (1):

$g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}$

g₂ = 11 m/s²