Halley's comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.586 A. U. and its greatest distance from the Sun being 35.5 A. U. (1 A. U. = the Earth-Sun distance). The comet's speed at closest approach is 49 km/s. What is its speed when it is farthest from the Sun, assuming that its angular momentum about the Sun is conserved?

Question

Spencer Montgomery

Physics

9 February, 19:55

Halley's comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.586 A. U. and its greatest distance from the Sun being 35.5 A. U. (1 A. U. = the Earth-Sun distance). The comet's speed at closest approach is 49 km/s. What is its speed when it is farthest from the Sun, assuming that its angular momentum about the Sun is conserved?

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Answers (1)

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Alexus Myers · Answer 1

0.81 km/s

Explanation:

r₀ = distance at closest approach from sun = 0.586 AU

r = distance at greatest distance from sun = 35.5 AU

v₀ = speed of the comet at closest approach = 49 km/s

v = speed of the comet at greatest distance = ?

Using conservation of angular momentum

m v₀ r₀ = m v r

Inserting the values

(49) (0.586) = v (35.5)

v = 0.81 km/s