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`)

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 depth of 3200km is (Radius of the earth =6400km ) is

If g on the surface of the earth is 9.8ms^(-2) , its value of a depth of 3200 km, (Radius of the earth =6400 km) is

The acceleration due to gravity at a depth d below the surface of the earth is 20% of its value at the surface. What is the value of d if the radius of the earth=6400km ?

If the value of g at the surface of the earth is 9.8 m//sec^(2) , then the value of g at a place 480 km above the surface of the earth will be (Radius of the earth is 6400 km)

The ratio of the orbital speeds of two satellites of the earth if the satellite are at heights 6400 km and 19200km (Raduis of the earth= 6400 km )

Compare the weights of a body when it is (i) 200 km above the surface of the Earth and (ii) 200 km below the surface of Earth. Radius of the Earth is 6400 km .

If the weight of a body is 100 N, on the surface of the earth then its binding energy on the surface of the earth is, (R_(E)=6400km)

Acceleration due to gravity is ‘ g ’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km )