Hailey\'s comet: The comet Hailey does NOT rotates around the Sun in a circular
ID: 1589930 • Letter: H
Question
Hailey's comet: The comet Hailey does NOT rotates around the Sun in a circular orbit. Instead it traverses an elliptical orbit (see figure). Point A shows the furthest distance from the sun, and Point B shows the closest distance to the sun. POINT A: the speed of Hailey's comet is 900.0 m/s. R^a = 5.24 x 10^12 m For this problem keep 4 sig figs for all values. Find the total energy at A. Find the acceleration at A. Find the speed at B. Find the acceleration at B. Find the work done on the comet from A to B, so half way around. Find the work done on the comet for the return trip, so from B back to A.Explanation / Answer
a) Gravitational PE = - GMm/d
= - (6.67 x 10^-11 x 1.989 x 10^30 x 2.2 x 10^14 ) / (5.24 x 10^12)
= - 5.57 x 10^21 J
KE = mv^2 / 2 = 2.2 x 10^14 x 900^2 /2 = 8.91 x 10^19
total energy = PE + KE = -5.48 x 10^22 J
b) at A.
F = ma
GM m / Ra^2 = ma
a = 1.327 x 10^20 / (5.24 x 10^12) = 4.83 x 10^-6 m/s^2
c) using energy conservation,
PE + KE at A = PE + KE at B
-5.48 x 10^22 J = [-(6.67 x 10^-11 x 1.989 x 10^30 x 2.2 x 10^14)/(8.83 x 10^10)] + mv^2 /2
v = 49931.97 m/s
d) using F = ma
(1.327 x 10^20 x m ) / (8.83 x 10^10)^2 = ma
a = 0.017 m/s^2
e) work done = - change in PE
= - [( - 5.57 x 10^21) - (3.31 x 10^23)] = 3.25 x 10^23 J
f) workd done = - 3.25 x 10^23 J
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