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Chapter 6: Problem 44

The orbit of Halley's comet around the Sun is a long thin ellipse. At its aphelion (point farthest from the Sun), the comet is $5.3 \times 10^{12} \mathrm{m}\( from the Sun and moves with a speed of \)10.0 \mathrm{km} / \mathrm{s} .$ What is the comet's speed at its perihelion (closest approach to the Sun) where its distance from the Sun is \(8.9 \times 10^{10} \mathrm{m} ?\)

Short Answer

Expert verified
Answer: The speed of Halley's comet at its perihelion is approximately \(5.95 \times 10^4 \mathrm{m/s}\).

Step by step solution

01

Write down the given information

We are given the following information: - Distance from the Sun at aphelion: \(r_a = 5.3 \times 10^{12} \mathrm{m}\) - Speed of the comet at aphelion: \(v_a = 10.0 \mathrm{km/s} = 10^4 \mathrm{m/s}\) - Distance from the Sun at perihelion: \(r_p = 8.9 \times 10^{10} \mathrm{m}\) We need to find the speed of the comet at perihelion (\(v_p\)).
02

Apply the conservation of angular momentum

As the angular momentum is conserved during the comet's movement around the Sun, we can write the following equation: \(\displaystyle L_{a}=L_{p}\) where \(L_a\) and \(L_p\) are the angular momenta at aphelion and perihelion, respectively.
03

Calculate the angular momenta at aphelion and perihelion

The equation for angular momentum is: \(\displaystyle L=mv \times r\) where \(m\) is the mass of the comet, \(v\) is its speed, and \(r\) is its distance from the Sun. Using this equation, we can write the conservation of angular momentum expression as follows: \(\displaystyle m v_{a} r_{a}=m v_{p} r_{p}\) Since the mass of the comet (\(m\)) does not change, we can simplify the equation to: \(\displaystyle v_{a} r_{a}=v_{p} r_{p}\)
04

Solve for the speed at perihelion

Using the given values and the equation from step 3, we have: \(\displaystyle (10^4 \mathrm{m/s}) (5.3 \times 10^{12}\mathrm{m}) =v_{p} (8.9 \times 10^{10} \mathrm{m})\) Now, solve for \(v_p\): \(\displaystyle v_{p} =\frac{(10^4 \mathrm{m/s}) (5.3 \times 10^{12}\mathrm{m})}{(8.9 \times 10^{10} \mathrm{m})}\) \(\displaystyle v_{p} \approx 5.95 \times 10^4 \mathrm{m/s}\)
05

Write down the final answer

The speed of Halley's comet at its perihelion is approximately \(5.95 \times 10^4 \mathrm{m/s}\).

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