Numerical Problem No. 5.8 (Gravitation) Physics SSC - I (Class 9)

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 = `\frac {g}{4}` = `\frac {10 ms⁻² }{4}`= 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 = `\frac {G m Mₑ}{(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 = `\frac {G m Mₑ}{(Rₑ + h)^2 }`

m will cancel from both sides hence 

gₕ = `\frac {G Mₑ}{(Rₑ + h)^2 }`

or

(Rₑ + h)² = `\frac {G Mₑ}{gₕ }`

taking square root on both sides

`\sqrt {(Rₑ + h)^2}` = `\sqrt \frac {G Mₑ}{gₕ }`

(Rₑ + h) = `\sqrt \frac {G Mₑ}{gₕ }`

h = `\sqrt \frac {G Mₑ}{gₕ }` - Rₑ

By putting values

h = `\sqrt \frac {6.67 ｘ 10⁻¹¹ N m² kg⁻² ｘ 6 ｘ 10²⁴ kg}{0.25 m s⁻²}` - 6.4 ｘ 10⁶ m

h = `\sqrt \frac {6.67 ｘ 6ｘ 10⁻¹¹⁺²⁴ m²}{0.25}` - 6.4 ｘ 10⁶ m

h = `\sqrt \frac {17.79ｘ 10¹³ m²}{1}` - 6.4 ｘ 10⁶ m

h = `\sqrt \frac {177.9ｘ 10¹² m²}{1}` - 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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