Question

If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet. The gravitational acceleration on Earth is 9.8 m/s 2 and the value of the universal gravitational constant is 6.67259 × 10−11 N · m2 /kg2 . Answer in units of m/s 2 .

Answer 1

Answer:

$8.1 m/s^2$

Explanation:

The strength of the gravitational field at the surface of a planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

$g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2$

For the unknown planet,

$M_X = 0.231 M_E\\R_X = 0.528 R_E$

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

$g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E$

And substituting g = 9.8 m/s^2,

$g_X = 0.829(9.8)=8.1 m/s^2$