A satellite orbits the Earth with a time period of 24 h at an orbital radius R . Another satellite orbits the Earth with a time period of 3 h . The orbital radius of the second satellite is:

Step-by-step solution

According to Kepler's third law of planetary motion, the square of the time period of a satellite is directly proportional to the cube of its orbital radius. T^2 R^3 For the two satellites, we can write the ratio as: ( R₂ R₁ )^3 = ( T₂ T₁ )^2 Given that T₁ = 24 h , R₁ = R , and T₂ = 3 h : ( R₂ R )^3 = ( 3 24 )^2 ( R₂ R )^3 = ( 1 8 )^2 = 1 64 Taking the cube root on both sides: R₂ R = [3] 1 64 = 1 4 R₂ = R 4 Answer: R 4