Calculation of true thickness of the bed

CSIR-NET 2015

65. Consider two planets A and B with radii 2r and r, respectively. Let their distances form the Sun be d and 2d , respectively. The solar constants for A (Fsa) and B (Fsb) are related by

A) $F_{sa} =\frac{1}{4}F_{sb}$ 
                            
B) $F_{sa} =F_{sb}$

C) $F_{sa} =4 F_{sb}$                                                       
D) $F_{sa} =2 F_{sb}$

Solution:

Solar constant $(S)=\frac{E\times R^2}{r^2}$ ------------------------(1)

R- radius of the Sun

r- radius of earth’s orbit around the Sun

( Sun- planets distance)

_{$F_{SA}=\frac{E\times R^2}{r_{SA}^2}$} ------------------(2)

r_SA – Sun- planet A distance

_{$F_{SB}=\frac{E\times R^2}{r_{SB}^2}$----------------------(3)}

r_SB- Sun-planet B distance

Divide equation (2) by (3)

$\bf \frac{F_{SA}}{F_{SB}}=\frac{\frac{E\times R^2}{r_{SA}^2}}{\frac{E\times R^2}{r_{SB}^2}}$

$\bf \frac{F_{SA}}{F_{SB}}=\frac{\frac{1}{r_{SA}^2}}{\frac{1}{r_{SB}^2}}$

Given in above problem r_SA= d

                                     r_SB= 2d

$\bf \frac{F_{SA}}{F_{SB}}=\frac{\frac{1}{d^2}}{\frac{1}{(2d)^2}}$

$\bf \frac{F_{SA}}{F_{SB}}=\frac{\frac{1}{d^2}}{\frac{1}{4d^2}}$

F_SA= 4 F_SB

Therefore Solar constant relation between planet A 

and planet B is given above.