/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 The orbit of Halley's comet arou... [FREE SOLUTION] | 91影视

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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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Most popular questions from this chapter

(a) Calculate the change in potential energy of \(1 \mathrm{kg}\) of water as it passes over Niagara Falls (a vertical descent of \(50 \mathrm{m}\) ). (b) \(\mathrm{At}\) what rate is gravitational potential energy lost by the water of the Niagara River? The rate of flow is $5.5 \times 10^{6} \mathrm{kg} / \mathrm{s} .\( (c) If \)10 \%$ of this energy can be converted into electric energy, how many households would the electricity supply? (An average household uses an average electrical power of about \(1 \mathrm{kW} .\) )
Show that 1 kilowatt-hour (kW-h) is equal to 3.6 MJ.
Human feet and legs store elastic energy when walking or running. They are not nearly as efficient at doing so as kangaroo legs, but the effect is significant nonetheless. If not for the storage of elastic energy, a \(70-\mathrm{kg}\) man running at \(4 \mathrm{m} / \mathrm{s}\) would lose about \(100 \mathrm{J}\) of mechanical energy each time he sets down a foot. Some of this energy is stored as elastic energy in the Achilles tendon and in the arch of the foot; the elastic energy is then converted back into the kinetic and gravitational potential energy of the leg, reducing the expenditure of metabolic energy. If the maximum tension in the Achilles tendon when the foot is set down is \(4.7 \mathrm{kN}\) and the tendon's spring constant is $350 \mathrm{kN} / \mathrm{m},$ calculate how far the tendon stretches and how much elastic energy is stored in it.
The bungee jumper of Example 6.4 made a jump into the Gorge du Verdon in southern France from a platform \(182 \mathrm{m}\) above the bottom of the gorge. The jumper weighed \(780 \mathrm{N}\) and came within \(68 \mathrm{m}\) of the bottom of the gorge. The cord's unstretched length is \(30.0 \mathrm{m}\) (a) Assuming that the bungee cord follows Hooke's law when it stretches, find its spring constant. [Hint: The cord does not begin to stretch until the jumper has fallen \(30.0 \mathrm{m} .]\) (b) At what speed is the jumper falling when he reaches a height of \(92 \mathrm{m}\) above the bottom of the gorge?
Two springs with spring constants \(k_{1}\) and \(k_{2}\) are connected in parallel. (a) What is the effective spring constant of the combination? (b) If a hanging object attached to the combination is displaced by \(2.0 \mathrm{cm}\) from the relaxed position, what is the potential energy stored in the spring for \(k_{1}=5.0 \mathrm{N} / \mathrm{cm}\) and $k_{2}=3.0 \mathrm{N} / \mathrm{cm} ?$ [See Problem \(83(\mathrm{b}) .]\)
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