Answer:
v = 6295.55 m/s
Explanation:
Given that,
The radius of a planet, [tex]r=5.32\times 10^6\ m[/tex]
The free fall acceleration of the planet, a = 7.45 m/s²
We need to find the tangential speed of a person standing at the equator.
Also, the centripetal acceleration was equal to the gravitational acceleration at the equator.
We know that,
Centri[etal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar}\\\\v=\sqrt{5.32\times10^6\times 7.45}\\\\v=6295.55\ m/s[/tex]
So, the tangential speed of the person is equal to 6295.55 m/s.
