Gravitation Tes...

The ratio of acceleration due to gravity at a depth h below the surface of earth and at a height h above the surface of earth for h < < R (where R is radius of earth)

The linear velocity of a body at the equator of the earth will be about $$($$Radius of earth at equator $$ =6.4 \times 10^{6} \mathrm{m} ) $$

The acceleration due to gravity becomes $$(\frac{g}{2})$$  where g=acceleration due to gravity on the surface of the earth at a height equal to

Who among the following did first give the experimental value of $$G$$?

At what height from the surface of earth the gravitation potential and the value of g are $$ -5.4 \times 10^7 J/kg $$ and $$ 6.0 m/s^2 $$ respectively?
take the radius of earth as 6400 km:

To double the orbital speed V and halve the angular velocity $$\omega $$, The centripetal acceleration of a revolving body:

The height above the surface of the earth where acceleration due to gravity is $$1/64$$ of its value at surface of the earth is approximately

If the radius of a planet shrinks by 2% while its mass remaining the unchanged, the acceleration due to gravity on the planet's surface would 

The ratio of acceleration due to gravity at a depth $$h$$ below the surface of earth and at a height $$h$$ above the earth surface of earth (h << radius of earth ) is 

The value of  $$g$$  will be  $$1\%$$  of its value at the surface of earth at a height of  $$\left( R _ { e } = 6400 k m \right).$$