Chapter 7, Problem 52A

a.

To determine

Rate of area swept out by Earth in Sun’s orbit.

Answer to Problem 52A

Rate of area swept out by Earth in Sun’s orbit is $2.3\times {10}^{15}{\text{m}}^2/\text{s}$ .

Explanation of Solution

Given:

Rate of area swept is measured in ${\text{m}}^{\text{2}}\text{/s}$

Formula used:

Area swept is given as $\frac{\pi r^2}{T}$

Calculation:

The distance between earth and Sun is $r=1.52\times {10}^{11}\text{m}$

Time period of revolution is

  $\begin{array}{c}T=365.25\text{days}\\ \text{=3}\text{.156}\times {\text{10}}^7\text{s}\end{array}$

So, area swept is given as $\frac{\pi r^2}{T}$

  $\begin{array}{c}A=\frac{3.14\times {\left(1.52\times {10}^{11}\right)}^2}{3.156\times {10}^7}\\ =2.3\times {10}^{15}{\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, rate of area swept out by Earth in Sun’s orbit is $2.3\times {10}^{15}{\text{m}}^2/\text{s}$ .

b.

To determine

Rate of area swept out by Moon in Earth’s orbit.

Answer to Problem 52A

Rate of area swept out by Moon in Earth’s orbit is $2\times {10}^{11}{\text{m}}^2/\text{s}$ .

Explanation of Solution

Given:

Average distance between Moon and Earth is $r=3.9\times {10}^8\text{m}$

Period of Moon is 27. 33 days

So, $T=2.36\times {10}^6\text{s}$

Formula used:

Area swept is given as $\frac{\pi r^2}{T}$

Calculation:

Area swept is,

  $\begin{array}{c}A=\frac{3.14\times {\left(3.9\times {10}^8\right)}^2}{2.36\times {10}^6}\\ =2\times {10}^{11}{\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, rate of area swept out by Moon in Earth’s orbit is $2\times {10}^{11}{\text{m}}^2/\text{s}$ .