Q69PE - College Physics (Urone)

Question

The Sun produces energy at a rate of \(4.00{\rm{ }} \times {\rm{ }}{10^{26}}\,W\) by the fusion of hydrogen. 

(a) How many kilograms of hydrogen undergo fusion each second? 

(b) If the Sun is \(90.0{\rm{ }}\% \)hydrogen and half of this can undergo fusion before the Sun changes character, how long could it produce energy at its current rate?

 (c) How many kilograms of mass is the Sun losing per second?

 (d) What draction of its mass will it have lost in the time found in part (b)?

Step-by-Step Solution

Answer

1. The amount of hydrogen that undergoes fusion reaction each second is obtained as:\(6.35{\rm{ }} \times {\rm{ }}{10^{11}}{\rm{ }}kg\).
2. The time fusion reaction will take to continue is obtained as:\(4.47{\rm{ }} \times {\rm{ }}{10^{10}}{\rm{ }}yr\).
3. The weight of mass sun is losing per second is obtained as:\(4.44{\rm{ }} \times {\rm{ }}{10^9}\,kg\).
4. The draction of mass that has lost in time found in part (b) is obtained as:\(3.14{\rm{ }} \times {\rm{ }}{10^{ - 3}}\).

Step 1: Define Special Relativity

The special theory of relativity, sometimes known as special relativity, is a physical theory that describes how space and time interact. Theoretically, this is known as STR theory.

Step 2: Evaluating the fusion sun undergoes each second

1. In fusion reaction of sun, there are four protons that fuses together to form the alpha particle.

The mass of a proton is of the value:\({m_p}{c^2}{\rm{ }} = {\rm{ }}938.3{\rm{ }}MeV\).

The mass alpha particle then is:

\(\begin{aligned}{m_\alpha }{c^2}{\rm{ }} &= {\rm{ }}4.001{\rm{ }}au\\ & = {\rm{ }}3726.9{\rm{ }}MeV\end{aligned}\)

So , the mass difference in the reaction is then obtained as:

\(\begin{aligned}\Delta m{c^2}{\rm{ }} & = {\rm{ }}4{m_p} - {m_\alpha }\\ & = {\rm{ }}4(938.3{\rm{ }}MeV){\rm{ }} - {\rm{ }}(3726.9{\rm{ }}MeV)\\ & = {\rm{ }}26.3{\rm{ }}MeV\\ & = {\rm{ }}4.21{\rm{ }} \times {\rm{ }}{10^{12}}{\rm{ }}J\end{aligned}\)

One \(H\) atom now has a mass of \(1.67{\rm{ }} \times {\rm{ }}{10^{ - 27}}{\rm{ }}kg\). As a result, the total amount of hydrogen required in one reaction is \({m_1}{\rm{ }} = {\rm{ }}6.68{\rm{ }} \times {\rm{ }}{10^{ - 27}}{\rm{ }}kg\). In one second, the sun produces \(E{\rm{ }} = {\rm{ }}4.00{\rm{ }} \times {\rm{ }}{10^{26}}{\rm{ }}J\) of energy. 

As a result, the total number of reactions that occur in a second is:

\(\begin{aligned}n{\rm{ }} & = {\rm{ }}\frac{{4.00{\rm{ }} \times {\rm{ }}{{10}^{26}}{\rm{ }}J}}{{4.21{\rm{ }} \times {\rm{ }}{{10}^{12}}{\rm{ }}J}}\\ & = {\rm{ }}9.50{\rm{ }} \times {\rm{ }}{10^{37}}\end{aligned}\)

Then , the total mass of hydrogen taking part in this reaction is obtained as:

\(\begin{aligned}{m_t}{\rm{ }} & = {\rm{ }}(9.50{\rm{ }} \times {\rm{ }}{10^{37}})(6.68{\rm{ }} \times {\rm{ }}{10^{ - 27}}{\rm{ }}kg)\\ & = {\rm{ }}6.35{\rm{ }} \times {\rm{ }}{10^{11}}{\rm{ }}kg\end{aligned}\)

Therefore, the amount of hydrogen undergo fusion reaction each second is:\(6.35{\rm{ }} \times {\rm{ }}{10^{11}}{\rm{ }}kg\).

Step 3: Evaluating how long will it take to produce the energy?

b. The total mass of sun is:\({m_s} = {\rm{ }}1.99{\rm{ }} \times {\rm{ }}{10^{30}}{\rm{ }}kg\).

The total mass of hydrogen in sun is obtained as:

\(\begin{aligned}{m_h} & = (90\% ){m_s}\\ & = (0.90)(1.99{\rm{ }} \times {\rm{ }}{10^{30}}{\rm{ }}kg)\\&  = 1.79{\rm{ }} \times {\rm{ }}{10^{30}}{\rm{ }}kg\end{aligned}\)

So, then the total amount of hydrogen that will take part in fusion reaction is:

\(\begin{aligned}{m_{hf}} & = \frac{1}{2}(1.79{\rm{ }} \times {\rm{ }}{10^{30}}{\rm{ }}kg)\\ & = 8.95{\rm{ }} \times {\rm{ }}{10^{29}}{\rm{ }}kg\end{aligned}\)    

So, the total time in which the fusion reaction can continue is:

\(\begin{aligned}t & = \frac{{8.95{\rm{ }} \times {\rm{ }}{{10}^{29}}{\rm{ }}kg}}{{6.35{\rm{ }} \times {\rm{ }}{{10}^{11}}{\rm{ }}kg}}\\ & = 1.41{\rm{ }} \times {\rm{ }}{10^{18}}{\rm{ }}s\\ & = {\rm{ }}4.47{\rm{ }} \times {\rm{ }}{10^{10}}{\rm{ }}yr\end{aligned}\)

Therefore, the fusion reaction will continue for:\(4.47{\rm{ }} \times {\rm{ }}{10^{10}}{\rm{ }}yr\).

Step 4: Evaluating the weight of mass sun is losing per second

c. The sun losing mass per second is obtained as:

\(\begin{aligned}\Delta {m_t} & = \frac{{4.00{\rm{ }} \times {\rm{ }}{{10}^{26}}{\rm{ }}J}}{{{{(3{\rm{ }} \times {\rm{ }}{{10}^8}{\rm{ }}m/s)}^2}}}\\ & = 4.44{\rm{ }} \times {\rm{ }}{10^9}{\rm{ }}kg\end{aligned}\)

Therefore, the mass that sun is losing per second is: \(4.44{\rm{ }} \times {\rm{ }}{10^9}\,kg\).

Step 5: Evaluating the draction of mass that has been lost in time found in the second part

So in the value of\(t{\rm{ }} = {\rm{ }}1.41{\rm{ }} \times {\rm{ }}{10^{18}}{\rm{ }}s\) the sun will loose:

\(\begin{aligned}\Delta {m_{t1}} & = \Delta {m_t}t\\ & = (4.44{\rm{ }} \times {\rm{ }}{10^9}{\rm{ }}kg)(1.41{\rm{ }} \times {\rm{ }}{10^{18}}{\rm{ }}s)\\ & = 6.26{\rm{ }} \times {\rm{ }}{10^{27}}\end{aligned}\)

Then, the total draction of mass used is obtained as:

\(\frac{{6.26{\rm{ }} \times {\rm{ }}{{10}^{27}}}}{{1.99{\rm{ }} \times {\rm{ }}{{10}^{30}}{\rm{ }}kg}} = 3.14{\rm{ }} \times {\rm{ }}{10^{ - 3}}\)

Therefore, the sun must have lost \(3.14{\rm{ }} \times {\rm{ }}{10^{ - 3}}\) draction of its mass.