Gravitation Tes...

The height at which the weight of a body becomes $$\dfrac {1}{16}th$$, its weight on the surface of earth (radius $$R$$), is

At what height above the surface of earth acceleration due to gravity becomes $$1 \ mm/s^2$$?
[Take $$R$$ as radius of earth and acceleration due to gravity at the surface of earth $$g = 10 m/s^2$$]

If $$'R'$$ is the radius of the earth, then the height at which the weight of a body becomes $$\dfrac{1}{4}^{th}$$ of its weight on the surface of the earth is:

A lift is descending with a constant velocity $$\ 'V'$$, A man in the lift drops a coin experiences an acceleration towards the floor equal to:

Two planet have radii in the ratio $$1:2$$ and densities in the ratio $$1:3$$. The value of the ratio between acceleration due to gravity on the first planet to that of the second planet is:

A body weighs $$900$$N on the earth. Find its weight on a planet whose density is $$\dfrac{1}{3}^{st}$$ the density of earth and radius is $$\dfrac{1}{4}^{th}$$ that of the earth.

The depth at which the value of acceleration due to gravity becomes $${\dfrac{1}{n}}$$ times the value at the surface is (R be the radius of the earth):

A particle when thrown. moves such that it passes from same height at $$2$$ and $$10 s$$, the height is:

The escape velocity of an object from the earth depends upon the mass of earth $$(M)$$, its mean density $$(\rho)$$, its radius $$(R)$$ and gravitational constant $$(G)$$, thus the formula for escape velocity is:

If the gravitational potential energy of two point masses infinitely away is taken to be zero then gravitational potential energy of a galaxy is