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πR²

   = 1400 × 4π × (1.5 × 10¹¹)²
   = 1400 × 4π × (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,

\[∆ m = \frac{E}{c^2}\]

\[\Rightarrow ∆ m = \frac{Pt}{c^2}\]

\[ \Rightarrow \frac{∆ m}{t} = \frac{P}{c^2}\]

\[ = \frac{1400 \times 4\pi \times 2 . 25 \times {10}^{22}}{9 \times {10}^{16}}\]

\[ = \left( \frac{1400 \times 4\pi \times 2 . 25}{9} \right) \times {10}^6 \]

\[ = 4 . 4 \times {10}^9 kg/s\]

So, Sun is losing its mass at the rate of \[4 . 4 \times {10}^9 kg/s .\]

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

2 × 10³⁰ kg disintegrates in t' = \[\frac{2 \times {10}^{30}}{4 . 4 \times {10}^9} s\]

\[\Rightarrow t' = \left( \frac{1 \times {10}^{21}}{2 . 2 \times 365 \times 24 \times 3600} \right)\]

\[ = 1 . 44 \times {10}^{- 8} \times {10}^{21} \]

\[ = 1 . 44 \times {10}^{13} y\]