Gravitation

The acceleration due to gravity at height $$h$$ above the earth if $$h << \mathrm{R}$$ (Radius of earth) is given by

The weight of a body on the earth is $$400 \mathrm{~N}$$. Then weight of the body when taken to a depth half of the radius of the earth will be:

The weight of a body on the surface of the earth is $$100 \mathrm{~N}$$. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

The mass of a spaceship is $$1000$$ $$kg.$$ It is to be launched from the earth's surface out into free space. The value of $$g$$ and $$R$$ (radius of earth ) are $$10\,m/{s^2}$$ and $$6400$$ $$km$$ respectively. The required energy for this work will be:

If $\mathrm{R}$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $\mathrm{h}=2 R$ from the surface of earth will be, :

The acceleration due to gravity on the surface of earth is $$\mathrm{g}$$. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be :

At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as $$R$$.)

A $$90 \mathrm{~kg}$$ body placed at $$2 \mathrm{R}$$ distance from surface of earth experiences gravitational pull of : ($$\mathrm{R}=$$ Radius of earth, $$\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}$$)

Assuming the earth to be a sphere of uniform mass density, a body weighed $$300 \mathrm{~N}$$ on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?

The escape velocity of a body depends upon mass as