Calculate the value of g at a height of 3600 km above the surface of the Earth. Ans. (4.0 ms⁻²) 


Given:  

Height above the Earth's Surface = h = 3600 km = 3.6 ｘ 10³ｘ 10³ m = 3.6 ｘ 10⁶ m

∴ Mass of the Earth = Mₑ = 6 ｘ 10²⁴ kg
∴ Radius of the Earth = Rₑ = 6.4 ｘ 10⁶ m
∴ Gravitational Constant = G = 6.67 ｘ 10⁻¹¹N m² kg⁻²


To Find:
Value of  Gravitational acceleration at height 3600 km above the Earth's surface = g =  ? m s⁻²



Solution:   

Let a body of mass m is placed at height h above the surface of the Earth So The distance between the centres of the body and the Earth = r = (Rₑ + h)

According to the Law of Gravitation's, we have 

F = GmMₑ(Rₑ+h)2   -----eqn (1)

But according to the 2nd law of Newton

F = W = mg ----eqn (2)

By comparing these two Eqns 1 & 2 we have 

mg = GmMₑ(Rₑ+h)2

m will cancel from both sides hence  

g = GMₑ(Rₑ+h)2

putting values

g = 6.67ｘ10⁻¹¹Nm²kg⁻²ｘ6ｘ10²⁴ kg(3.6 ｘ 10⁶m+ 6.4ｘ 10⁶ m)2

g = 6.67ｘ10⁻¹¹Nm²kg⁻²ｘ6ｘ10²⁴ kg(10ｘ 10⁶ m)2

g = 6.67ｘ10⁻¹¹Nm²Kg⁻²ｘ6ｘ10²⁴ kg(100ｘ 10¹² m2}

g = 6.67ｘ6100ｘ 10⁻¹¹⁺²⁴⁻¹² m s⁻²

by simplifying we get

g = 0.4003ｘ 10¹ m s⁻²

g = 4.003ｘ 10⁻¹ ｘ 10¹ m s⁻²

g = 4.003ｘ 10⁻¹⁺¹ m s⁻²

or
g = 4.003ｘ10⁰ m s⁻²

g = 4.0 m s⁻²

Thus, the gravitational acceleration g on the above surface of Earth at height 3600 Km is 4.0 ms⁻²

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