Answer:
Gravity of hypothetical planet is [tex]\dfrac{g}{27}[/tex].
Explanation:
Let the mass of Earth be 'M' and radius be 'R'.
Given:
Mass of the hypothetical planet (m) = one-third of Earth's mass = [tex]\frac{M}{3}[/tex]
Radius of hypothetical planet (r) = 3 times Earth's radius = [tex]3R[/tex]
We know that, acceleration due to gravity of a planet of mass 'M' and radius 'R' is given as:
[tex]g=\dfrac{GM}{R^2}[/tex]
Now, the above is the value of 'g' for Earth.
Now, acceleration due to gravity of hypothetical planet is given as:
[tex]g_{hyp}=\dfrac{Gm}{r^2}\\\\g_{hyp}=\dfrac{G\times\frac{M}{3}}{(3R)^2}\\\\g_{hyp}=\dfrac{GM}{3\times 9R^2}\\\\g_{hyp}=\frac{1}{27}(\frac{GM}{R^2})=\frac{1}{27}\times g[/tex]
So, the hypothetical planet is 1/27 times of the gravity of the Earth.
