Two stars of masses 3 $$ \times $$ 10³¹ kg each, and at distance 2 $$ \times $$ 10¹¹ m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is - (Take Gravitational constant; G = 6.67 $$ \times $$ 10^–11 Nm² kg^–2)

A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 $$ \times $$ 10³ km) is E₁ and kinetic energy required for the satellite to be in a circular orbit at this height is E₂. The value of h for which E₁ and E₂ are equal, is

Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence :