   Chapter 8, Problem 66P

Chapter
Section
Textbook Problem

Hailey’s comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.59 AU and its greatest distance being 35 AU (1 AU is the Earth-Sun distance). If the comet’s speed at closest approach is 54 km/s, what is its speed when it is farthest from the Sun? You may neglect any change in the comet’s mass and assume that its angular momentum about the Sun is conserved.

To determine
The speed of comet Halley when it is farthest from the Sun.

Explanation

Given info: The closest distance of the comet from the Sun is 0.59AU , the farthest distance of the comet from the Sun is 35AU , and the speed of the comet when it is at closest distance to the Sun is 54km/s .

Explanation: The angular momentum of the comet is conserved and when it approach aphelion and perihelia, the conservation of angular momentum of the comet is Laphelion=Lperihelion and it can be modified as (mra2)ωa=(mrp2)ωp . In general, the angular speed of the comet is related as ω=v/r . Using all these expressions in the expression of conservation of angular momentum, the speed of the comet at a distance farthest from the Sun is determined.

The formula for the speed of the comet at farthest distance from the Sun is,

va=(rpra)vp

• rp is closest distance of the comet from the Sun

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