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QPage 458: Q22: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q22 Solution: The energy from the sun reaches just outside the earth's…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently calculate the sun's total power by multiplying the solar constant ($1400 \text{ W m}^{-2}$) by the cross-sectional area of the Earth ($\pi R_{\text{Earth}}^2$) or simply by $1 \text{ m}^2$, ignoring that the radiation is distributed over the entire surface area of a sphere with radius equal to the Sun-Earth distance. Always use the full spherical area $4\pi r^2$ to determine the total luminosity before applying $P = \dot{m}c^2$.

The energy from the sun reaches just outside the earth's atmosphere at a rate of

1400 \text{ W m}^{-2}

. The distance between the sun and the earth is

1.5 \times 10^{11} \text{ 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 \times 10^{30} \text{ kg}

.

Step-by-Step Verified Solution

**Answer:** (a) Rate of mass loss =

4.67 \times 10^9

kg/s (b) Sun will last

1.36 \times 10^{13}

years. **Step-by-Step Explanation:** (a) Power received at earth per m_ = 1400 W/m_. Total power of sun =

1400 \times 4\pi(1.5 \times 10^{11})^2 = 3.96 \times 10^{26}

W. Rate of mass loss:

dm/dt = P/c^2 = 3.96 \times 10^{26}/(9 \times 10^{16}) = 4.4 \times 10^9

kg/s. (b) Time =

M/(dm/dt) = 2 \times 10^{30}/(4.4 \times 10^9) = 4.5 \times 10^{20}

s =

1.4 \times 10^{13}

years.

All Questions in this Chapter

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