Calculate the change in the value of g at altitude `45^(@)`. Take radius of earth `R = 6.37 xx 1-0^(3)` km`.

Calculate the value of acceleration due to gravity at a place of latitude 45^(@) . Radius of the earth = 6.38 xx 10^(3) km .

An artificial satellite circles around the earth at a height of 2,200 km. Calculate its orbital velocity and period of revolution. Take, Radius of earth = 6.37 xx 10^(3) km Mass of earth = 6 xx 10^(24) kg G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)

What is the value of g at a height 8848 m above sea level. Given g on the surface of the earth is 9.8 ms^(-2) . Mean radius of the earth = 6.37 xx 10^(6) m .

If the earth were made of iron of relative density 7.87, what would be the value of acceleration due to gravity on its surface ? Take, Radius of earth = 6.37 xx 10^(6) m , G = 6.67 xx 10^(11) Nm^(2) kg^(-2)

At what height from the surface of earth will the value of g becomes 40% from the value at the surface of earth. Take radius of the earth = 6.4 xx 10^(6) m .

At what altitude, the acceleration due to gravity reduces to one fourth of its value as that on the surface of the earth ? Take radius of earth as 6.4 xx 10^(6) m , g on the surface of the earth as 9.8 ms^(-2) .

At what altitude, the acceleration due to gravity reduces to one fourth of its value as that on the surface of the earth ? Take radius of earth as 6.4 xx 10^(6) m , g on the surface of the earth as 9.8 ms^(-2) .

An artificial satellite completes a circle around earth in 120 minutes. Calculate the height of the satellite above the earth's surface. Take, radius of earth = 6,400 km g = 9.8 m//s^(2)

At what altitude , the acceleration due to gravity reduces to half of its value as that on the surface of the earth ? Take radius of earth as 6 . 4 xx 10^(6) m , g on the surface of the earth as 9.8 m//s ^(2) .