Chapter 16, Problem 39A

(a)

To determine

To Find:The distance to which light travel in 13s from one orbit of Jupiter to the next orbit.

Answer to Problem 39A

The distances travelled by the light in 13s is $3.9\times {10}^6 \text{km}$ .

Explanation of Solution

Given:

The average time is 13 s .

Formula:

Distance travelled by the light in a particular time is equal to the product of the speed of light and the time.

  $d=ct$  ...... (1)

Here c is the speed of light.

Calculation:

Substitute $3\times {10}^8 \text{m}/\text{s}$ for c and $13\text{s}$ for t in equation (1)

  $\begin{array}{c}\text{d}=13\times 3\times {10}^8 \text{m}/\text{s}\text{}\\ \text{d}=39\times {10}^{8 } \text{m}\left(\frac{1 \text{km}}{{10}^3 \text{m}}\right)\text{}\\ \text{d}= 3.9\times {10}^6 \text{km}\end{array}$

Conclusion:

Hence, the required distance travelled by the light in 13s is $3.9\times {10}^6 \text{km}$ .

(b)

To determine

To Find:The speed of earth in kilometers per second.

Answer to Problem 39A

  $28\text{ km/s}$

Explanation of Solution

Given:

Time taken by the orbit is 42.5 h

Formula:

Speed of the earth to travel a distance can be obtained as follows:

  $V_{Earth}=\frac{d}{t}$  ...... (1)

Here, d is the distance travelled by the earth.

Calculation:

Substitute $3.9\times {10}^{6 } \text{km}$ for d and 42.5 h for t in equation (1)

  $\begin{array}{c}{\text{V}}_{Earth}=\frac{3.9\times {10}^6 \text{km}}{42.5 \text{h}}\\ \text{}{\text{V}}_{Earth}=0.1\times {10}^6\text{ km/h}\left(\frac{\text{1 hr}}{\text{60 min}}\right)\left(\frac{1 \min }{60 \mathrm{s}}\right)\text{}\\ {\text{V}}_{Earth}=28\text{ km/s}\end{array}$

Conclusion:

Hence, the required speed of earth is $28\text{ km/s}$ .

(c)

To determine

To Find: The speed of earth in the orbit using the orbital radius.

Answer to Problem 39A

  $30 \text{km/s}$

Explanation of Solution

Given:

Radius of the orbit is $r=1.5\times {10}^8\text{km}$

Period of the orbit is $t=$ 1 year.

Formula:

The orbital speed of the earth which is equal to the circumference of the earth’s orbit divided by the orbital period of the earth

  $v=\frac{2\pi r}{t}$  ...... (1)

Where, r is the radius of the earth and t is the earth orbital period.

Calculation:

  $\begin{array}{c}v=\frac{6.28\times 1.5\times {10}^8}{1}\\ \text{}v=9.42\times {10}^8 \text{km/y}\left(\frac{\text{1}}{\text{365d}}\right)\left(\frac{\text{1}}{\text{24hr}}\right)\left(\frac{\text{1}}{\text{60min}}\right)\left(\frac{\text{1}}{\text{60s}}\right)\\ \text{}v=30 \text{km/s}\end{array}$

Conclusion:

Hence, the required orbital speed of earth is $30 \text{km/s}$ .