The time period of a satellite of earth is 5 hours If the separation between the earth and the earth and the satellite is increased to 4 times the previous value, the new time period will become(a) 10 hours(b) 80 hours(c) 40 hours(d) 20 hours

(c) According to Kepler’s law of planetary motion T² ? R³ ? T₂ = T₁ (R₂/R_1­)^3/2 = 5 x [4R/R]^3/2 = 5 x 2³ = 40 hour

Time period = total distance/velocityT=2πr/√grT`=2π(4r)/√g(4r)T`=4(2πr)/2√grT`=2(2πr)/√grT`=2TT`=2×5T`=10 hours(option A)

Time period = total distance/velocity T =2πr/√gr T`=2π(4r)/√g(4r) T`=4(2πr)/2√gr. T`=2(2πr)/√gr T`=2T T`=2×5T`=10 hours(option A)