At what altitude the value of g would become one fourth than on the surface of the Earth? Ans. (One Earth's radius)


Given:  

Gravitational  acceleration on the surface of the Earth = g =  10 ms⁻² 

Value of  Gravitational  acceleration at height h = gₕ = g4g4 = 10ms⁻²4= 0.25 m s⁻²

∴ 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:
height above the Earth = h = ? meters



Solution:   

Let a body of mass m is placed at height h where the value of gₕ becomes one-fourth than on 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 = mgₕ ----eqn (2)

By comparing these two Eqns 1 & 2 we have 

mgₕ = GmMₑ(Rₑ+h)2

m will cancel from both sides hence 

gₕ = GMₑ(Rₑ+h)2

or

(Rₑ + h)² = GMₑgₕ

taking square root on both sides

√(Rₑ+h)2 = √GMₑgₕ

(Rₑ + h) = √GMₑgₕ

h = √GMₑgₕ - Rₑ

By putting values

h = √6.67ｘ10⁻¹¹Nm²kg⁻²ｘ6ｘ10²⁴kg0.25ms⁻² - 6.4 ｘ 10⁶ m

h = √6.67 ｘ6ｘ10⁻¹¹⁺²⁴ m²0.25 - 6.4 ｘ 10⁶ m

h = √17.79ｘ10¹³ m²1 - 6.4 ｘ 10⁶ m

h = √177.9ｘ10¹² m²1 - 6.4 ｘ 10⁶ m

h = 13.3 ｘ 10⁶ m  - 6.4 ｘ 10⁶ m

or

h = 6.9 ｘ 10⁶ m

At this height of one Earth's Radius i.e. 6.9 ｘ 10⁶ m) above  the surface of the earth the value of gravitational acceleration becomes one-fourth than on the surface of the Earth

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