Answer:
The free fall acceleration on the surface of this planet is 4.35 m/s²
Explanation:
Given that,
Mass of planet [tex]M_{p}=\dfrac{1}{9}M_{e}[/tex]
Radius of planet [tex]R_{p}=\dfrac{1}{2}R__{e}[/tex]
We need to calculate the free fall acceleration on the surface of this planet
Using formula of gravity
[tex]g_{p}=\dfrac{GM_{p}}{R_{p}}[/tex]
Where, Â [tex]M_{p}[/tex] = mass of planet
[tex]R_{p}[/tex] = radius of plane
Put the value into the formula
[tex]g_{p}=\dfrac{G\times\dfrac{M_{e}}{9}}{\dfrac{R_{e}^2}{4}}[/tex]
[tex]g_{p}=\dfrac{4}{9}\times\dfrac{GM_{e}}{R_{e}^2}[/tex]
[tex]g_{p}=\dfrac{4\times9.8}{9}[/tex]
[tex]g_{p}=4.35\ m/s^2[/tex]
Hence, The free fall acceleration on the surface of this planet is 4.35 m/s²
