Chapter 31, Problem 61GP

(a)

To determine

Total power output of the Sun.

Answer to Problem 61GP

The total power output is $\mathrm{P}=3.96\times {10}^{26}\text{W}$

Explanation of Solution

Given:

Rate of energy, $\mathrm{I}=1400{\text{ W/m}}^2$

Formula used:

Power output can be given as:

  $\mathrm{P}=\mathrm{I}\left(4\mathrm{\pi }{\mathrm{r}}^2\right)$

Calculation:

Distance between the Earth and the Sun is $\mathrm{r}=1.5\times {10}^{11}\text{m}$

Substituting all the values, in the above formula:

  $\begin{array}{l}\mathrm{P}=\left(1400{\text{ W/m}}^2\right)\left(4\right)\left(3.14\right){\left(1.5\times {10}^{11}\text{m}\right)}^2\\ \mathrm{P}=3.96\times {10}^{26}\text{W}\end{array}$

Conclusion:

Hence, the total power output is $\mathrm{P}=3.96\times {10}^{26}\text{W}$

(b)

To determine

Number of protons consumed per second in the reaction.

Answer to Problem 61GP

The number of protons per second is $\mathrm{N}=3.706\times {10}^{38}$

Explanation of Solution

Given:

Rate of energy, $\mathrm{I}=1400{\text{ W/m}}^2$

Calculation:

Energy released for the reaction in eq 31-7,

  $\begin{array}{l}\mathrm{E}=26.7\text{ MeV}\\ \mathrm{E}=26.7\times {10}^6\times 1.6\times {10}^{-19}\text{J}\\ \mathrm{E}=160.2\times {10}^{-13}\text{J}\end{array}$

So, the number of protons per second is:

  $\begin{array}{l}\mathrm{N}=\frac{\left(3.96\times {10}^{26}\text{W}\right)\left(4\right)}{160.2\times {10}^{-13}\text{J}}\\ \mathrm{N}=3.706\times {10}^{38}\text{protons/s}\end{array}$

Conclusion:

Thus, the number of protons per second is $\mathrm{N}=3.706\times {10}^{38}$

(c)

To determine

Time taken by the Sun to glow at its present rate.

Answer to Problem 61GP

The time taken by the Sun to glow at its present rate will be $\mathrm{t}={10}^{11}\text{yrs}$

Explanation of Solution

Given:

Rate of energy, $\mathrm{I}=1400{\text{ W/m}}^2$

Mass of Sun, $\mathrm{m}=2.0\times {10}^{30}\text{kg}$

Calculation:

The time taken can be calculated as follows:

  $\begin{array}{l}\mathrm{t}=\frac{\left(2.0\times {10}^{30}\text{kg}\right)}{\left(1.67\times {10}^{-27}\text{kg}\right)\left(3.706\times {10}^{38}\text{protons/s}\right)\left(3.156\times {10}^7\text{s}\right)}\\ \mathrm{t}={10}^{11}\text{yrs}\end{array}$

Conclusion:

Hence, the time taken by the Sun to glow at its present rate will be $\mathrm{t}={10}^{11}\text{yrs}$