A geostationary satellite is orbiting the earth at a height of 4R above the surface of the earth, where R is the radius of the earth. Another satellite is orbiting the earth at a height 1.5R from the surface of the earth with periodic time 'T' in hour. The value of 'T' in hour is

Step-by-step solution

The orbital radius of the geostationary satellite is r₁ = R + 4R = 5R . The time period of a geostationary satellite is T₁ = 24 hours. The orbital radius of the second satellite is r₂ = R + 1.5R = 2.5R . Let the time period of the second satellite be T₂ = T . According to Kepler's third law, the square of the time period is proportional to the cube of the orbital radius: T^2 r^3 ( T₂ T₁ )^2 = ( r₂ r₁ )^3 ( T 24 )^2 = ( 2.5R 5R )^3 ( T 24 )^2 = ( 1 2 )^3 = 1 8 T 24 = 1 2 2 T = 24 2 2 = 6 2 hours Answer: 6 2