Gravitation Tes...

 The variation of acceleration due to gravity $$g$$ with distance $$d$$ from centre of the earth is best represented by ($$R=$$Earths radius):

 The height at which the acceleration due to gravity becomes $$g/9$$ (where $$g=the$$ acceleration due to gravity on the surface of the earth) in terms of $$R$$, the radius of the earth, is

The fractional change in the value of free-fall acceleration $$'g'$$ for a particle when it is lifted from the surface to an elevation $$h. (h < < R)$$ is

At what depth below the surface of the earth acceleration due to gravity $$'g'$$ will be half of its value at $$1600\ km$$ above the surface of the earth?

Weight of a body on earth's surface is $$W$$. At a depth half way to the centre of the earth, it will be (assuming uniform density in earth).

A body of mass 'm' is raised from the surface of the earth to a height 'nR' (R- radius of earth). Magnitude of the change in the gravitational potential energy of the body is (g-acceleration due to gravity on the surface of earth)

The radius of a planet is $$4$$ times the radius of the earth. The time period of revolution of the planet will be:

In a double star system one of mass $$m_{1}$$ and another of mass $$m_{2}$$ with a separation $$d$$ rotate about their common centre of mass. Then rate of sweeps of area of star of mass $$m_{1}$$ to star of mass $$m_{2}$$ about their common centre of mass is

If potential at the surface of a planet is taken as zero, the potential at infinity will be $$(M$$ and $$R$$ are mass and radius of the planet).

Weight of a body decreases by $$1.5$$%, when it is raised to a height $$h$$ above the surface of the earth. When the same body is taken to same depth $$h$$ in a mine, its weight will show