The escape velocity from the earth’s surface is 11 km/sec. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of the escape velocity from this planet would be

If g is the acceleration due to gravity of the earth’s surface the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

An earth’s satellite of mass m revolves in a circular orbit at a height h from the surface g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

The mean radius of the earth is R, its angular speed on its own axis is ω and the acceleration due to gravity at earth’s surface is g. The cube of the radius of the orbit of a geo-stationary satellite will be

The radii of the earth and the moon are in the ratio 10 : 1 while acceleration due to gravity on the earth’s surface and moon’s surface are in the ratio 6 : 1. The ratio of escape velocities from earth’s surface to that of moon surface is

The escape velocity from earth is 11.2 km per sec. If a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is

If the gravitational force between two objects were proportional to 1/R (and not as 1/R²) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

What would be the duration of the year if the distance between the earth and the sun gets doubled?

If escape velocity from the earth’s surface is 11.2 km/sec. then escape velocity from a planet of mass same as that of earth but radius one fourth as that of earth is

If ‘r’ represents the radius of the orbit of a satellite of mass ‘m’ moving round a planet of mass ‘M’, the velocity of the satellite is given by