We know from Stefan Boltzmann Law P = e sigma T^4 A where e is the emmissivity ;

If Pa = 5 W/m^2 if the star B also at the same distance, Pb = 5/62.43 = 0.08 W/m^2 but when star B in only half as far away, Pb = 0.08*4 = 0.32 W/m^2

ID: 1600884 • Letter: W

Question

We know from Stefan Boltzmann Law

P = e sigma T^4 A

where e is the emmissivity ; sigma is the Boltzmann constant ; T is the surface temp ; A is the Area

Pa = e sigma (2000)^4 [ 4 pi (40d/2)^2]

Pb = e sigma (4500)^4 [ 4 pi (d/4)^2 ]

Pa/Pb = e sigma (2000)^4 [ 4 pi (40d/2)^2]/ e sigma (4500)^4 [ 4 pi (d/4)^2 ]

Pa/Pb = (2000)^4 x 40^2/(4500)^2 = 62.43

Hence, Pa/Pb = 62.43 ; star a is brighter than star b , by a factor of 62.43.

if star A above is magnitude 5 as seen from earth what is the magnitude of star b if it only half as far away?

If Pa = 5 W/m^2

if the star B also at the same distance, Pb = 5/62.43

= 0.08 W/m^2

but when star B in only half as far away,

Pb = 0.08*4

= 0.32 W/m^2 <<<<<<<------------Answer

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