A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth's surface.
Answers
Answered by
0
Answer:
Explanation:
a) V = √GM/r + h = √gr^2/r + h = √9.8 * (6400 * 10^3)^2/10^6 * (6.4 + 2) = 6.9 * 10^3 m/s = 6.9 km/s b) K.E. = (1/2) mv^2 = (1/2) 1000 * (47.6 * 10^+6) = 2.38 * 10^10 J c) P.E. = GMm/-(R + h) = - 6.67 * 10^-11 * 6 * 10^24 * 10^3/(6400 + 2000) * 10^3 = - 40 * 10^-13/8400 = - 4.76 * 10^10 J d) T = 2π(r + h)/V = 2 * 3.14 * 8400 * 10^3/6.9 * 10^3 = 76.6 * 10^2 sec = 2.1 hour
Similar questions