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Physical World

Question

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

0.5 gms^-2

0.99 gms^-2

1.01 gms^-2

0.90 gms^-2

Solution

0.99 gms^-2

For height 'h' above the earth's surface

$g = g (1 - \frac{2 h}{R_{e}}) = g (1 - \frac{64}{6400}) (since R_{e} = 6400 km) = g (1 - 0.01) 0.99 {gms}^{- 2}$