Answer:
Acceleration of gravity on the planet is [tex]\frac{g}{27}[/tex]
Explanation:
The acceleration of gravity at the surface of a planet is given by
[tex]g=\frac{GM}{R^2}[/tex]
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
For the Earth, we have
[tex]g_E = \frac{GM_E}{R_E^2}[/tex]
where [tex]M_E[/tex] is the Earth mass and [tex]R_E[/tex] is the Earth's radius.
For the planet in this problem, we have
[tex]M_X = \frac{1}{3}M_E[/tex] (mass is 1/3 of the Earth)
[tex]R_X=3R_E[/tex] (radius is 3 times that of Earth)
Therefore, its acceleration of gravity is
[tex]g_X = \frac{GM_X}{R_X^2}=\frac{G(\frac{1}{3}M_E)}{(3R_E)^2}=\frac{1}{27}(\frac{GM_E}{R_E^2})=\frac{g_E}{27}[/tex]
