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## Q. 16.4

Estimate the mean temperature of the sun by assuming its protons behave as a gas.

Strategy As gas clouds collected together through the gravitational interaction, the decrease in the gravitational potential energy was accompanied by an increase in the kinetic energy of the particles. We can estimate the mean temperature of the sun by setting the change in the kinetic energy equal to the negative of the change in potential energy, $\Delta( K . E .)=-\Delta( P . E .)$. In other words, the total energy, K.E. + P.E., is constant. We use the kinetic theory of gases to relate the mean velocity to temperature.

## Verified Solution

We assume the sun is a uniform sphere of mass M and radius R. Its self-potential energy can be calculated as (see Problem 28)

\begin{aligned}\text { P.E. } &=-\frac{3}{5} \frac{G M^{2}}{R} \\&=-\frac{3}{5} \frac{\left(6.67 \times 10^{-11} N \cdot m ^{2} / kg ^{2}\right)\left(1.99 \times 10^{30} kg \right)^{2}}{6.96 \times 10^{8} m } \\&=-2.28 \times 10^{41} J\end{aligned}

The kinetic energy of the particles within the sun is then $2.28 \times 10^{41}$ J. If we assume the sun is made entirely of protons, the number of protons in the sun is $N=M / m_{p}$, where M is the mass of the sun. We write the kinetic energy of the sun as N protons each having speed v.

$\frac{1}{2} N m_{p} v^{2}=\frac{1}{2} \frac{M}{m_{p}} m_{p} v^{2}=2.28 \times 10^{41} J$

From this relation we can determine the proton velocity.

\begin{aligned}v^{2} &=\frac{2\left(2.28 \times 10^{41} J \right)}{M}=\frac{2\left(2.28 \times 10^{41} J \right)}{1.99 \times 10^{30} kg } \\&=2.3 \times 10^{11} m ^{2} / s ^{2} \\v &=4.8 \times 10^{5} m / s\end{aligned}

Using the kinetic theory of gases relation for the rms speed of a gas particle at temperature T, we can determine the sun’s mean temperature from Equation (9.20).

$v=\sqrt{\frac{3 k T}{m}}$ (9.20)

We determine T to be

$T=\frac{m v^{2}}{3 k}$

For a proton, we determine the temperature to be

$T=\frac{\left(1.67 \times 10^{-27} kg \right)\left(2.3 \times 10^{11} m ^{2} / s ^{2}\right)}{3\left(1.38 \times 10^{-23} J / K \right)}=9 \times 10^{6} K$

This is likely to be lower than the actual mean temperature of the sun due to the energy input from nuclear fusion.