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
B
0.99gms2
C
1.01gms2
D
0.90gms2

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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