# The acceleration due to gravity a height 1/20th of the radius of earth above the earth surface is 9 m/sq.sec, Its value at a point at an equal distance below the surface of earth is what

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Let us consider the height 1/20 th of Radius to be small compared to the Radius.

h /R = 1/20

acceleration due to gravity at an altitude h: g' = g (1 - 2 h / R) = g * 18/20

g = 9 * 20 / 18 = 10 m/sec/sec

acceleration due to gravity at depth h : g" = g (1 - h/R) = g * ( 1 - 1/20)

g" = g * 19/20

= 10 * 19/20 = 9.5 m/sec/sec

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we know gravity reduces by half the amount, from the surface of Earth at depth = altitude.... g above surface is 9 m/sec square, which is 1 less than on sthe surface... So at same distance from surface under the surface, the gravity will reduce by 0.50...

so it will be 10 - 0.50 = 9.50 m/sec/sec

h /R = 1/20

acceleration due to gravity at an altitude h: g' = g (1 - 2 h / R) = g * 18/20

g = 9 * 20 / 18 = 10 m/sec/sec

acceleration due to gravity at depth h : g" = g (1 - h/R) = g * ( 1 - 1/20)

g" = g * 19/20

= 10 * 19/20 = 9.5 m/sec/sec

==================================

we know gravity reduces by half the amount, from the surface of Earth at depth = altitude.... g above surface is 9 m/sec square, which is 1 less than on sthe surface... So at same distance from surface under the surface, the gravity will reduce by 0.50...

so it will be 10 - 0.50 = 9.50 m/sec/sec