CBSE-XI-Physics
47: The Special Theory of Relativity
- #22The 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.digAnsr: bAns : 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}}``
`` \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\,}``
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\,}``
`` \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``
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