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 =
Radius of new planet (r) = 6 times Earth's radius =
We know that, acceleration due to gravity of a planet of mass 'M' and radius 'R' is given as:
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 ](https://tex.z-dn.net/?f=g_%7Bnew%7D%3D%5Cdfrac%7BGm%7D%7Br%5E2%7D%5C%5C%5C%5Cg_%7Bnew%7D%3D%5Cdfrac%7BG%5Ctimes%5Cfrac%7BM%7D%7B6%7D%7D%7B%286R%29%5E2%7D%5C%5C%5C%5Cg_%7Bnew%7D%3D%5Cdfrac%7BGM%7D%7B6%5Ctimes%2036R%5E2%7D%5C%5C%5C%5Cg_%7Bnew%7D%3D%5Cfrac%7B1%7D%7B216%7D%28%5Cfrac%7BGM%7D%7BR%5E2%7D%29%3D%5Cfrac%7B1%7D%7B216%7D%5Ctimes%20g%0A)
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](https://tex.z-dn.net/?f=g_%7Bnew%7D%3D%5Cfrac%7B9.8%7D%7B216%7D%3D0.045%5C%20m%2Fs%5E2)
Therefore, the free fall acceleration on the surface of this planet is 0.045 m/s².