Question

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

Short answer

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

Step-by-step solution

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 = $\mathbf{3}.\mathbf{84}\times {\mathbf{10}}^6\mathbf{m}$

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.

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

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:

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

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 $\mathbf{1}\mathbf{\%}$.