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Question

Acceleration due to gravity at earth's surface is '$$g$$' m/s$$^2$$. Find the effective value of acceleration due to gravity at a height of $$32 km$$ from sea level :($$R_e = 6400 km$$).


A
0.5gms2
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B
0.99gms2
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C
1.01gms2
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D
0.90gms2
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Solution

The correct option is C $$0.99 g ms^{2}$$
Acceleration due to gravity at a height h from the earth surface,  $$g'=g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$$
Given  $$h= 32  km         \quad                   R=  6400   km$$
Using bionomial expansion, $$(1+x)^n= 1+ nx$$               for  $$x<< 1$$

Thus as $$h<< R$$   $$\implies g'=g\bigg[1-\dfrac{2h}{R}\bigg]$$

$$ g'=g\bigg[1-\dfrac{2(32)}{6400}\bigg]$$  
 $$\implies g'= .99  g         m/s^2$$

Physics

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