Find the value of gravitational acceleration g on the earth's surface. G=6.67xx10^(-11) Nm^(2)//kg^(2) Earth's fluid...

The mass of the earth is 6.0 xx 10^(24) kg and its radius is 6.4 xx 10^(6) m . How much work will be done in taking a 10 kg body from the surface of the Earth to infinity ? What will be the gravitational potential energy of the body on the Earth's surafce ? G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

The radius of the earth is 6.37xx10^(6) m its mass 5.98xx10^(24) kg determine the gravitational potential on the surface of the earth Given G=6.67 xx10^(-11) N-m^(2)//khg^(2)

If the value of universal gravitational constant is 6.67xx10^(11) Nm^2 kg^(-2), then find its value in CGS system.

Taking moon's period of revolution about earth as 20 days and neglecting the effect of the sun and the effect other planet on its motion, calculate its distance from the earth. Given G=6.67xx10^(-11)Nm^(-2)kg^(-2) and mass of the earth =6xx10^(24)kg .

The value of gravitational acceleration 'g' at a height 'h' above the earth's surface is (g)/(4) , then (R = __________ ) (where R = radius of the earth)

Calculate the force of attraction between two balls each of mass 10kg when their centres are 10 cm apart. The value of gravitational constant G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)

The radius of the moon is 1.7 xx 10^(6) m and its mass is 7.35 xx 10^(22) kg . What is the acceleration due to gravity on the surface of the moon ? Given G = 6.67 xx 10^(-11) Nm^(2) kg^(-2).

The mean radius of a planet is 6.67xx10^(3) km. The acceleration due to gravity on its surface is 10 m//s^(2) . If G = 6.67xx10^(-11)Nm^(2)//kg^(2) , then the mass of the planet will be [R=6.67xx10^(6)m]