Acceleration due to gravity on earth is g . Find the effective value of acceleration due to gravity at height 32km from sea level
R_e= 6400 km

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Solution

Variation of acceleration due to gravity with height is given by the relation :
g ' = g R^2/( R + h)^2 ........(1)
where g is the value of acceleration due to gravity on the earth's surface = 9.8 m/s2
R is the radius of the earth = 6400 km
h is the height above the surface of the earth = 32 km
g'=9.8(6400)^2/(6400+32)^2
g'=401408000/41370624
g'=9.702m/s^2

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Similar questions

The acceleration due to gravity becomes g/2 (g= acceleration due to gravity on the surface of earth) at a height equal to

64 k m

32 k m

= 6400 k m

= 9.8 m s^{- 2}

4 %

[R_{e} = 6400 k m]

g_{1} = g [\frac{R}{R + h}]^{2}

R_{e} = 6.4 \times 10^{6} m]

1600 k m

= 6400 k m

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