# An object is dropped from a height of 4.05*10^6 m above the surface of the Earth. What is its initial acceleration? The acceleration of gravity is 9.81 m/s^2 and the radius of the earth is...

An object is dropped from a height of 4.05*10^6 m above the surface of the Earth. What is its initial acceleration? The acceleration of gravity is 9.81 m/s^2 and the radius of the earth is 6.37*10^6 m.

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The gravitational force of attraction between two objects with a mass M1 and M2 is and which are separated by a distance r is equal to F = G*M1*M2/r^2 where G is a constant.

Let the mass of the object that is dropped be m. On the surface of the Earth, the force of attraction is G*m*Me/(6.37*10^6)^2. And the acceleration of the object is G*Me/(6.37*10^6)^2 = 9.81 m/s^2

When the object is dropped from a height of 4.05*10^6, its distance from the center of the Earth is 10.42*10^6 m. The acceleration of the object at this height is a = G*Me/(10.42*10^6)^2.

Substituting G*Me = 9.81*(6.37*10^6)^2,

a = 9.81*(6.37*10^6)^2/(10.42*10^6)^2

=> 9.81*(6.37^2/10.42^2)

=> 3.66 m/s^2

The acceleration of the object when it is dropped from a height of 4.05*10^6 is 3.66 m/s^2

The great distance from the earth reduces the acceleration of gravity.

## Acceleration of gravity in m/s^2 = G m(earth)/r²

G = 6.673e-11 m³/kgs²

m(earth) = 5.97e24 kg

r is the distance in meters

earth radius = 6,371,000 m

a = Gm/r²a = (6.373e-11)(5.97e24) / (6,371,000+4,050,000)² = (6.373e-11)(5.97e24) / (10,421,000)² = (approx) 3.5 m/s^2