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At what depth below the surface does the acceleration due to gravity becomes 70% of its value on the surface of the earth?

Important Questions on Gravitation

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At what depth from earth's surface does the acceleration due to gravity becomes 14 times that of its value at surface ?
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If earth is assumed to be a sphere of uniform density then plot a graph between acceleration due to gravity (g) and distance from the centre of earth.

 

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The gravitational acceleration on the surface of earth is g. Find the increase in potential energy in lifting an object of mass m to a height equal to the radius of earth.
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In a certain region of space gravitational field is given by I = – (K/r) (Where r is the distance from a fixed point and K is constant). Taking the reference point to be at r = r0 with V = V0. Find the potential at a distance r.
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Two masses of 102 kg and 103 kg are separated by 1 m distance. Find the gravitational potential at the mid point of the line joining them.
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The magnitude of the intensity of the gravitational field at a point situated at a distance 8000 km from the centre of earth is 6.0 N kg. The magnitude of the gravitational potential at that point in N m kg-1 will be:
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The gravitational field due to a mass distribution is, E=Kx3 in the x-direction. Here, K is a constant. Taking the gravitational potential to be zero at infinity, its value at a distance x is,
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A body of mass m is situated at distance 4Re above the earth's surface, where Re is the radius of earth how much minimum energy be given to the body so that it may escape :-