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Q10PE

Expert-verifiedFound in: Page 82

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 ${\mathbf{3}}{\mathbf{.}}{\mathbf{84}}{\mathbf{\times}}{{\mathbf{10}}}^{{6}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{m}}{\mathbf{\text{}}}\mathbf{\left(}\mathbf{1}\mathbf{\%}\mathbf{\right)}$** **?**

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

- 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 = $\mathbf{3}.\mathbf{84}\times {\mathbf{10}}^{6}\text{\hspace{0.17em}}\mathbf{m}$

**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:

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

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

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