A man stands on vertical tower of height 20 cm. Calculate the distance up to which he will be able to see on the surface of the earth. Neglect the height of the man. Take the radius of the earth `= 6400` km

If g on the surface of the earth is 9.8m//s^(2) , its value at a height of 6400 km is (Radius of the earth =6400 km)

If g on the surface of the earth is 9.8 m//s^2 , its value at a height of 6400km is (Radius of the earth =6400km )

At what height above the surface of earth the value of "g" decreases by 2 % [ radius of the earth is 6400 km ]

The height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (radius of the earth = 6400 km ) is

At what height above the surface of the earth will the acceleration due to gravity be 25% of its value on the surface of the earth ? Assume that the radius of the earth is 6400 km .

Find the velocity of a satellite at height 80 km from earth. If the radius of earth is 6400 km

Calculate the depth below the surface of the earth where the acceleration due to gravity is 2% of its value at the earth's surface. Take, Radius of earth = 6,400 km

Calculate the depth below the surface of the earth where acceleration due to gravity becomes half of its value at the surface of the earth . Radius of the earth = 6400 km.

A body is projected vertically upward from the surface of the earth with escape velocity. Calculate the time in which it will be at a height (measured from the surface of the arth) 8 time the radius of the earth (R). Acceleration due to gravity on the surface of the earth is g.