The acceleration due to gravity at height `R` above the surface of the earth is `g/4`. The periodic time of simple pendulum in an artifical satellie at this height will be :-

The acceleration due to gravity at a height h above the surface of the earth has the same value as that at depth d= ____below the earth surface.

Show that acceleration due to gravity at height h above the Earth’s surface is g_h = g ((R )/(R+h))^2

At what height above the surface of earth acceleration due to gravity will be (i) 4%

The acceleration due to gravity on the surface of moon is 1.7 ms^(-2) . What is the time period of a simple pendulum on the moon if its time period on the earth is 3.5 s ? (g on earth = 9.8 ms^(-2) )

The acceleration due to gravity at a height above the earth's surface is ……..than that at the same depth below the surface of the earth. [Fill in the blanks]

If g_(h) is the acceleration due to gravity at a height h above the earth's surface and R is the radius of the earth then, the critical velocity of a satellite revolving round the earth in a circular orbit at a height h is equal to

If the value of acceleration due to gravity at a height of 100 km from the surface of earth is g ',then at what depth ,value of acceleration due to gravity wioll again be g'?