CSIR-NET 2014 (119)

CSIR-NET 2014

119. If both the Sun-Earth distance and the temperature of the Sun's photo sphere decrease by 50% then the Solar constant.

A) decreases to 25% of its original value

B) increases to 400% of its original value

C) decrease to 50% of its original value

D) remains unchanged

Solution:

The energy falling per unit area = (E*R²)/r²

--------------------(1)

R- Radius of the Sun

r- Radius of Earth’s orbit around the Sun ( Sun -Earth distance)

S= (E*R²)/r²

but E = σ * T⁴ --------------------(2) (According to Stefan’s law)

S= (σ * T⁴*R²) / r²-----------------------(3)

T- temperature of the Sun

The relation between the temperature of the Sun (T)

and the Sun-Earth distance (r)

Solar constant (S) = T⁴/ r² -----------------(4)

From above problem, Sun-Earth distance and

temperature of the Sun decreases to 50%

So, S₁ = (T/2)⁴/ (r/2)²

S₁ = (T⁴/16) / (r²/4)

S₁ = T⁴ / 4*r²

S₁ = 1/4 *(T⁴/r²) ---------------(5)

Substituent eq (4) in eq (5)

Then, S₁ = 1/4 *(S)

Therefore Solar constant decreases 25% of its original value

GEOPHYSICS