When a falling meteor is at a distance above the Earth's surface of 2.98 times the Earth's radius, what is its free-fall acceleration caused by the gravitational force exerted on it

0.63 m/s²

Explanation:

Given that

Distance of meteor from earth's surface, r = 2.98R,

where R = radius of the Earth.

The force of the meteor due to the Earth is usually given by

F = GMM(a)/(R + r)² where

F = force exerted

G = gravitational constant

M = Mass of the earth

M(a) = mass of meteor

Remember that F = Ma, so applying that here, we have

acceleration of meteor = GM/(R + r)²

where r = 2.98R so that

acceleration of meteor = GM/(R + 2.98R)²

acceleration of meteor = GM/(3.98R)²

acceleration of meteor = GM / 15.84R²

It should be noted that GM/R² = 10, so

acceleration of meteor = 10/15.84

acceleration of meteor = 0.63 m/s²

Therefore as calculated, the acceleration of the meteor is 0.63 m/s²

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