Question

The energy from the sun reaches just outside the earth's atmosphere at a rate of 1400 W m⁻². The distance between the sun and the earth is 1.5 × 10¹¹ m.
(a) Calculate the rate which the sun is losing its mass.
(b) How long will the sun last assuming a constant decay at this rate? The present mass of the sun is 2 × 10³⁰ kg.

Answer 1

Given:
Intensity of energy from Sun, I = 1400 W/m²
Distance between Sun and Earth, R = 1.5 × 10¹¹ m

Power = Intensity × Area
P = 1400 × A
= 1400 × 4$\pi$R²
= 1400 × 4$\pi$ × (1.5 × 10¹¹)²
= 1400 × 4$\pi$ × (1.5)² × 10²²

Energy = Power × Time
Energy emitted in time t, E = Pt

Mass of Sun is used up to produce this amount of energy. Thus,
Loss in mass of Sun, $∆m=\frac{E}{c^2}$
$$\begin{gathered} \Rightarrow ∆m=\frac{Pt}{c^2} \\ \Rightarrow \frac{∆m}{t}=\frac{P}{c^2} \\ =\frac{1400\times 4\mathrm{\pi }\times 2.25\times {10}^{22}}{9\times {10}^{16}} \\ =\left(\frac{1400\times 4\mathrm{\pi }\times 2.25}{9}\right)\times {10}^6 \\ =4.4\times {10}^9\mathrm{kg}/\mathrm{s} \end{gathered}$$

So, Sun is losing its mass at the rate of $4.4\times {10}^9\mathrm{kg}/\mathrm{s}.$

(b) There is a loss of 4.4 × 10⁹ kg in 1 second. So,

2 × 10³⁰ kg disintegrates in t' = $\frac{2\times {10}^{30}}{4.4\times {10}^9}\mathrm{s}$
$$\begin{gathered} \Rightarrow t\text{'}=\left(\frac{1\times {10}^{21}}{2.2\times 365\times 24\times 3600}\right) \\ =1.44\times {10}^{-8}\times {10}^{21} \\ =1.44\times {10}^{13}y \end{gathered}$$