Question

If the rotation of a planet of radius 5.32 × 106 m and free-fall acceleration 7.45 m/s 2 increased to the point that the centripetal acceleration was equal to the gravitational acceleration at the equator, what would be the tangential speed of a person standing at the equator?

Answer 1

Answer:

v = 6295.55 m/s

Explanation:

Given that,

The radius of a planet, $r=5.32\times 10^6\ m$

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,

$a=\dfrac{v^2}{r}\\\\v=\sqrt{ar}\\\\v=\sqrt{5.32\times10^6\times 7.45}\\\\v=6295.55\ m/s$

So, the tangential speed of the person is equal to 6295.55 m/s.