Question

Two stars have the same luminosity, but Star A is 10.0 times farther away than Star B. How much fainter does Star A appear compared to Star B? times fainter

Answer 1

Answer:

B(A)100=B(B), the star A is 100 times fainter than star B.

Explanation:

Brightness of the star is defined by the formula,

$B=\frac{L}{4\pi d^{2}}$

Here, L is the luminosity and d is the distance.

For star A, the distance is 10d. The brightness of star A.

$B(A)=\frac{L}{4\pi (10d)^{2}}$

For star B, the distance is d. The brightness of star B.

$B(B)=\frac{L}{4\pi d^{2}}$

Now according to the question luminosity of two stars is equal.

Therefore,

$B(A){4\pi (10d)^{2}}=B(B){4\pi (d)^{2}}\\B(A)100=B(B)$

So, star B is 100 times brighter than star A.

Therefore the star A is 100 times fainter than star B.