Question

6. When Halley's comet is closest to the Sun, it is at a distance of 8.823x1010 m from the Sun and at a speed of 5.46x 104 m/s. Given the comet's mass is about 2.2x1014 kg and the Sun's mass is 2.00x1020 kg: (a) Find the total mechanical energy of Halley's comet (8 points) (b) Assume all the time Hallye's comet is subject to the Sun's gravitational pull only. As the comet moves further away from the Sun, what will happen to its speed? (3points) Explain why. Give the name of the law that applies. (3 points)

Answer 1

M elementof = k elementof + p elementof = m v^2/2 + (- G M_s m/r) M elementof = (2.2 times 10^14) [(5.46 times 10^4)/2^2 - (6.67 times 10^- 11) (2 times 10^30)/8.823 times 10^10] M elementof = 1.727 times 10^22 J when it moves further, its speed decreases T elementof = p elementof + k elementof = constant {Energy conservation} p elementof = G M_s m/r: r (increases) p elementof (increases because of, -ve sign) finally p elementof is maximum, p elementof, p elementof_max = 0} So k elementof increases . 1.727 times 10^22 = (2.2 times 10^14) v^2/2 is v = 1.25 times 10^4 m/s