With what velocity must a body be thrown from earth's surface so that it may reach a maximum height of `4R_(e)` above the Earth's surface ? (Radius of the Earth `R_(e)=6400 km,g=9.8 ms^(-2)`).

A projectile is launched vertically from earth's surface with an inittial velocity of 10 km/s. Calculate the height it will reach above the earth's surface. Take, Radius of earth = 6,400 km g = 9.8 m//s^(2)

At what height above the earth's surface the acceleration due to gravity will be 1/9 th of its value at the earth’s surface? Radius of earth is 6400 km.

At what height above the earth's surface the acceleration due to gravity will be 1/9 th of its value at the earth’s surface? Radius of earth is 6400 km.

At what height above the earth's surface the acceleration due to gravity will be 1/9 th of its value at the earth’s surface? Radius of earth is 6400 km.

The escape velocity of a body from earth's surface is v_e . The escape velocity of the same body from a height equal to 7R from earth's surface will be

A rocket is launched vetically from the surface of the earth with an initial velocity of 10 kms^(-1) . How for above the surface of the earth would it go ? Radius of the earth =6400 km and g=9.8 ms^(-2) .

At what height from the surface of earth will the value of g be reduced by 36% from the value on the surface? Take radius of earth R = 6400 km .

At what height from the surface of earth will the value of g be reduced by 36% from the value on the surface? Take radius of earth R = 6400 km .