What is the fractional decrease in the value of free-fall acceleration g for a particle when it is lifted from the surface to an elevation `h`? `(h lt lt R)`

If g denotes the value of acceleration due to gravity at a point distance r from the centre of earth of radius R . If r lt R , then

If the change in the value of g at a height h above the surface of the earth is same as at depth .x. below the surface of the earth, then (h lt lt R)

What is the weight felt by a person in a lift when it has free fall?

If R= radius of the earth and g= acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r lt R) from the centre of the earth is proportional to

Imagine earth to be a solid sphere of mass M and radius R. If the value of acceleration due to gravity at a depth d below earth's surface is same as its value at a height h above its surface and equal to (g)/(4) (where g is the value of acceleration due to gravity on the surface of earth), the ratio of (h)/(d) will be

Assume that the earth is a uniform sphere of radius R. If the acceleration due to gravity at a height h above the earth's surface is g_1 and at the same depth below the surface of the earth is g_2(h lt R) , then show that, g_2=g_1(1-h/R)(1+h/R)^2 .

The acceleration due to gravity is g at a point distant r from the centre of earth of radius R . If r lt R , then