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Found in: Page 82

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately ${\mathbf{4}}{\mathbf{\text{\hspace{0.17em}cm/year}}}$. Assuming this to be a constant rate, how many years will pass before the radius of the Moon’s orbit increases by ${\mathbf{3}}{\mathbf{.}}{\mathbf{84}}{\mathbf{×}}{{\mathbf{10}}}^{6}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{m}}{\mathbf{\text{}}}\mathbf{\left(}\mathbf{1}\mathbf{%}\mathbf{\right)}$ ?

$\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{7}\mathbf{\text{\hspace{0.17em}y}}$

See the step by step solution

## Step 1: Given Data

• Speed $\mathbf{=}\mathbf{4}\mathbf{\text{\hspace{0.17em}cm/y}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{\text{\hspace{0.17em}m/y}}$
• Total radius = $\mathbf{3}.\mathbf{84}×{\mathbf{10}}^{6}\text{\hspace{0.17em}}\mathbf{m}$

## Step 2: Time is taken to increase the 1% radius of the moon

Speed can be calculated as the distance that is traveled by an object in a time period.

${\mathbf{Speed}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{=}}\frac{\mathbf{Distance}}{\mathbf{Time}}$

It is a scalar quantity.

The time taken can be calculated as:

${\mathbf{Time}}{\mathbf{=}}\frac{\mathbf{Distance}}{\mathbf{Speed}}$

Substituting the values in the above expression, we get:

$\begin{array}{c}\mathrm{T}=\frac{3.84×{10}^{6}\text{\hspace{0.17em}m}}{0.04\text{\hspace{0.17em}m/y}}\\ =96,000,000\text{\hspace{0.17em}}\mathrm{y}\end{array}$

Hence,$\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{7}\mathbf{\text{\hspace{0.17em}y}}$ is needed to increase the radius of the moon by $\mathbf{1}\mathbf{%}$.