a) At the earth's surface, the radiant power received from the Sun normally is $1.3 \times 10^{3} \mathrm{~W}$ per square metre. The power radiated by the Sun is the same everywhere over the Sun's surface. If the Earth orbits at a distance of $1.5 \times 10^{11} \mathrm{~m}$ from the Sun, calculate the total energy radiated away by the Sun each second. (It may be useful to know that the surface area of a sphere is $4 \pi r^{2}$ ).
b) Although you may not have studied it yet, Einstein produced a famous equation relating mass and energy which we shall use, $E=m c^{2}$, where $E$ is energy in joules, $m$ is mass in $\mathrm{kg}, c$ is the velocity of light in a vacuum ( $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ). Using your answer to part (a), calculate the mass loss of the Sun due to the energy being radiated away each second.
c) If the mass of the Sun is $2 \times 10^{30} \mathrm{~kg}$, what is the percentage of the Sun's mass that is lost by radiation each year?
d) Assuming that this rate remains constant, what is the percentage loss of mass of the sun since it was formed, five thousand million years ago?
Show worked solution
The radiant power received at Earth is given as $P_{received} = 1.3 \times 10^3$ W/m$^2$ at a distance of $r = 1.5 \times 10^{11}$ m from the Sun.
This power is spread over a sphere centered on the Sun with radius equal to the Earth's orbital distance. The surface area of this sphere is:
$$A = 4\pi r^2 = 4\pi (1.5 \times 10^{11})^2$$
$$A = 4\pi \times 2.25 \times 10^{22} = 2.83 \times 10^{23} \text{ m}^2$$
The total power radiated by the Sun (energy per second) is:
$$P_{total} = P_{received} \times A$$
$$P_{total} = (1.3 \times 10^3) \times (4\pi \times (1.5 \times 10^{11})^2)$$
$$P_{total} = 1.3 \times 10^3 \times 4\pi \times 2.25 \times 10^{22}$$
$$P_{total} = 3.68 \times 10^{26} \text{ W} \approx 4 \times 10^{26} \text{ W}$$
b) Mass loss of the Sun per secondUsing Einstein's equation $E = mc^2$, we can find the mass equivalent of the radiated energy:
$$m = \frac{E}{c^2} = \frac{P_{total} \times t}{c^2}$$
For $t = 1$ second and $c = 3 \times 10^8$ m/s:
$$m = \frac{3.68 \times 10^{26}}{(3 \times 10^8)^2} = \frac{3.68 \times 10^{26}}{9 \times 10^{16}}$$
$$m = 4.09 \times 10^9 \text{ kg} \approx 4 \times 10^9 \text{ kg/s}$$
This is about 4 million tonnes per second!
c) Percentage of Sun's mass lost each yearMass loss per second: $4.09 \times 10^9$ kg/s
Mass loss per year (365 days): $$m_{year} = 4.09 \times 10^9 \times 60 \times 60 \times 24 \times 365$$
$$m_{year} = 4.09 \times 10^9 \times 31,536,000 = 1.29 \times 10^{17} \text{ kg/year}$$
Percentage of Sun's mass ($M_{Sun} = 2 \times 10^{30}$ kg): $$\text{Percentage} = \frac{1.29 \times 10^{17}}{2 \times 10^{30}} \times 100$$
$$\text{Percentage} = 6.45 \times 10^{-14} \% \approx 6 \times 10^{-14} \%$$
d) Percentage mass loss since formationAge of the Sun: 5000 million years $= 5 \times 10^9$ years
Total percentage loss: $$\text{Total loss} = 5 \times 10^9 \times 6.45 \times 10^{-14} \%$$
$$\text{Total loss} = 3.23 \times 10^{-4} \% \approx 3 \times 10^{-4} \%$$
Or in the original format: $3 \times 10^{-10} \%$ (using the rounded value from part c)
Note: The extremely small mass loss shows that the Sun will remain essentially unchanged for billions of years to come.