Answered: a planet has a moon which is 2.5x10^9 m…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter6: Motion In Two Dimensions

Section6.2: Circular Motion

Problem 13PP

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Question

a planet has a moon which is 2.5x10^9 m away from the planet, the moon makes a circular orbit around the planet with a period of 18 days, what is the centripetal acceleration in m/s^2 of the moon?

Expert Solution

Step 1

Given Data:

The distance between the moon and the planet is, $r = 2.5 \times 10^{9} m$

The time period of revolution in a circular orbit is, $t = 18 days$

The expression for the centripetal acceleration is given as,

$a_{c} = \frac{v^{2}}{r} . . . . . . . . . . . . . . . (1)$

Here, v is the orbital velocity and r is the distance.

The orbital velocity is given as,

$v = \frac{2 π r}{t} . . . . . . . . . . . . . (2)$

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