(a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of speed of earth’s rotation. (b) If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth = 6 × 1024 kg.

Sol. Angular speed f earth & the satellite will be same 2π/T base e = 2π/T base s Or 1/24 * 3600 = 1/2π√(R + h)^3/gR^2 or 12 l 3600 = 3.14 √(R + h)^3/gR^2 Or √(R + h)^3/gR^2 = (12 * 3600)^2/(3.14)^2 or (6400 + h)^3 + 10^9/9.8 * (6400)^2 * 10^6 = (12 * 3600)^2/(3.14)^2 Or (6400 + h)^3 * 10^9/6272 * 10^9 = 432 * 10^4 Or (6400 + h)^3 = 6272 * 432 * 10^4 Or 6400 + h = (6272 * 432 * 10^4)^1/3 Or h = (6272 * 432 * 10^4)^1/3 – 6400 = 42300 cm. b) Time taken from north pole to equator = (1/2) t = (1/2) * 6.28 √(43200 + 6400)^3/10 * (6400)^2 * 10^6 = 3.14 √(497)^3 * 10^6/(64)^2 * 10^11 = 3.14 √497 * 497 * 497/64 * 64 * 10^5 = 6 hour.