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 :

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n^th power of R. If the period of rotation of the particle is T, then :

Take the mean distance of the moon and the sun from the earth to be $$0.4 \times {10^6}$$ km and $$150 \times {10^6}$$ km respectively. Their masses are $$8 \times {10^{22}}$$ kg and $$2 \times {10^{30}}$$ kg respectively. The radius of the earth is $$6400$$ km. Let $$\Delta {F_1}$$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $$\Delta {F_2}$$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $${{\Delta {F_1}} \over {\Delta {F_2}}}$$ is :