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.
2 Answers | Add Yours
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
=> 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
a = Gm/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 = (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
We’ve answered 319,207 questions. We can answer yours, too.Ask a question