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Physics / Physics with MasteringPhysics 4 / Chapter 12 / Problem 82GP

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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Using the results from 54, find the angular momentum of

Chapter 12, Problem 82GP

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QUESTION:

Problem 82GP

Using the results from Problem 54, find the angular momentum of Halley's comet (a) at perihelion and (b) at aphelion. (Take the mass of Halley's comet to be 9.8 × 1014 kg.)

REFERENCE PROBLEM:

Halley’s Comet, which passes around the sun every 76 years has an elliptical orbit. When closest to the sun (perihelion) it is ar a distance of 8.823 X 1010 m and moves with a speed of 54.6 km/s. The greatest distance between Halley’s Comet and the Sun (aphelion is 6.152 x 1012 m.

Is the speed of Halley’s Comet greater than or less than 54.6 km/s when it is at aphelion? Explain.Calculate its speed at aphelion

Questions & Answers

QUESTION:

Problem 82GP

Using the results from Problem 54, find the angular momentum of Halley's comet (a) at perihelion and (b) at aphelion. (Take the mass of Halley's comet to be 9.8 × 1014 kg.)

REFERENCE PROBLEM:

Halley’s Comet, which passes around the sun every 76 years has an elliptical orbit. When closest to the sun (perihelion) it is ar a distance of 8.823 X 1010 m and moves with a speed of 54.6 km/s. The greatest distance between Halley’s Comet and the Sun (aphelion is 6.152 x 1012 m.

Is the speed of Halley’s Comet greater than or less than 54.6 km/s when it is at aphelion? Explain.Calculate its speed at aphelion

ANSWER:

Solution 82GP

Step 1 of 2

Part a

We are required to calculate the angular momentum of the comet at perihelion.

The distance of the comet from the sun at perihelion is $r{\ }_{p}=8.823\times{}1{0}^{10}$ m

The speed of the comet at this position $v{\ }_{p}=54.6$ km/s

$v{\ }_{p}=54600$ m/s

Mass of the comet $m=9.8\times{}1{0}^{14}$ kg

Therefore, the angular momentum of the comet is,

$L{\ }_{p}=mv{\ }_{p}r{\ }_{p}$

$L{\ }_{p}=9.8\times{}1{0}^{14}\times{}54600\times{}8.823\times{}1{0}^{10}$ kg $\cdot{}$ m2/s

$L{\ }_{p}=4.7\times{}1{0}^{6}\times{}1{0}^{24}$ kg $\cdot{}$ m2/s

$L{\ }_{p}=4.7\times{}1{0}^{30}$ kg $\cdot{}$ m2/s

Therefore, the angular momentum of the comet at perihelion is $4.7\times{}1{0}^{30}$ kg $\cdot{}$ m2/s.

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