Question

comet halley, discovered by edmond halley in 1758, has a very eccentric orbit. the aphelion of the orbit is 35.14 au and the perihelion is 0.59278 au. if the speed of the comet at the aphelion is 0.91 km/s. find the speed of the comet at the perihelion. note that 1 au = 1.496 × 10^{11} m.

Explanation:

Step1: Apply the law of conservation of angular momentum

The law of conservation of angular momentum states that \(L = mvr\) (where \(m\) is the mass of the comet, \(v\) is the speed, and \(r\) is the distance from the sun). Since angular momentum is conserved (\(L_{aphelion}=L_{perihelion}\)), we have \(m v_{a}r_{a}=m v_{p}r_{p}\). The mass \(m\) cancels out, so \(v_{p}=\frac{v_{a}r_{a}}{r_{p}}\).

Step2: Substitute the given values

We are given \(v_{a} = 0.91\space km/s\), \(r_{a}=35.14\space au\), and \(r_{p}=0.59278\space au\).

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Answer:

The speed of the comet at the perihelion is approximately \(54\space km/s\)