Question

A new planet has been discovered that has a mass one-sixth that of Earth and a radius that is six times that of Earth. Determine the free fall acceleration on the surface of this planet. Express your answer in the appropriate mks units.

Answer 1

Answer:

0.045 m/s²

Explanation:

Let the mass of Earth be 'M' and radius be 'R'.

Given:

Mass of the new planet (m) = one-sixth of Earth's mass = $\frac{M}{6}$

Radius of new planet (r) = 6 times Earth's radius = $6R$

We know that, acceleration due to gravity of a planet of mass 'M' and radius 'R' is given as:

$g=\dfrac{GM}{R^2}$

Now, this is acceleration due to gravity on Earth.

Now, acceleration due to gravity of new planet is given as:

$g_{new}=\dfrac{Gm}{r^2}\\\\g_{new}=\dfrac{G\times\frac{M}{6}}{(6R)^2}\\\\g_{new}=\dfrac{GM}{6\times 36R^2}\\\\g_{new}=\frac{1}{216}(\frac{GM}{R^2})=\frac{1}{216}\times g$

Now, the value of 'g' on Earth is approximately 9.8 m/s². So,

$g_{new}=\frac{9.8}{216}=0.045\ m/s^2$

Therefore, the free fall acceleration on the surface of this planet is 0.045 m/s².