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a meteor is falling towards the earth. if the mass and radius of the earth are 6×10^24 kg and 6.4×10^6 m respectively, find the height of the meteor where its acceleration due to gravity is 4m/s^2 ?
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Sagot :

Answer:

F = G M m / R^       force between m and earth where R >= radius of earth

a = F / m = G M / R^2        acceleration at radius R

am / ae = (Re / Rm)^2     acceleration of meteor to that of earth

am / ar = (4 / 9.8) = (Re / Rm)^2

Rm = (9.8 / 4)^1/2 Re

Rm = 1.56 Re   = 1.56 * 6.4E106 m = 10E6

10E6 - 6.4E6 = 3.6E6 m above surface of earth

(You only need mass of earth if you are calculating a at the surface of the earth)

Calculate g = G M / R^2 =  6.67E-11 * 6E24 / 6.4E6)^2  = 9.77 m/s^2

This is close to the value we used - 9.8 m/s^2

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