# 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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by santhoshkrish0302

2015-10-19T22:02:10+05:30

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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
thank u very much sir