Answer:
[tex]8.1 m/s^2[/tex]
Explanation:
The strength of the gravitational field at the surface of a planet is given by
[tex]g=\frac{GM}{R^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
For the Earth:
[tex]g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2[/tex]
For the unknown planet,
[tex]M_X = 0.231 M_E\\R_X = 0.528 R_E[/tex]
Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:
[tex]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[/tex]
And substituting g = 9.8 m/s^2,
[tex]g_X = 0.829(9.8)=8.1 m/s^2[/tex]
