The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value then its new time period will become

Step-by-step solution

Kepler's Third Law establishes T^2 r^3 for orbital motion. Substituting the given initial period T₁ = 24 hours and radius ratio r₂ r₁ = 1 4 into T₂^2 T₁^2 = ( r₂ r₁ )^3 yields: T₂^2 = (24)^2 ( 1 4 )^3 = 576 1 64 = 9 Taking the square root gives the new period T₂ = 3 hours.