Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is :

The radius of earth is 6400 km and the value of g is 10m/s² . If the weight of 5km body on the equator becomes zero, then the angular speed of earth will be :

The weight of a body at earth surface is 700 g wt. What will be its weight on a planet whose mass is 1/7 that of earth and radius half that of earth?

A satellite moves eastwards very near the surface of the earth in the equatorial plane of the earth with speed v₀. Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If R is the radius of the earth and ω be its angular speed about its own axis, then the difference in the two time period as observed on the earth will be approximately equal to :

The kinetic energy needed to project a body of mass m from the earth's surface (radius R) to infinity is :

Energy required to move a body of mass m from an orbit of radius 2R to 3R is :

If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will :

Suppose the gravitational force varies inversely as the n^th power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to :

If the radius of the earth is made three times, keeping its mass constant, then the weight of a body on earth's surface will be as compared to its previous value :

A particle of mass 10g is kept on the surface of a uniform sphere of mass 100kg and radius 10cm. Find the work to be done against the gravitational force between them, to take the particle far away from the sphere :

(You may take G = 6.67 × 10^-11 Nm²/kg²)