Question

Andromeda galaxy is the closest large galaxy and is visible to the naked eye. Estimate its brightness relative to the Sun, assuming it has luminosity\({\rm{1}}{{\rm{0}}^{{\rm{12}}}}\)times that of the Sun and lies\({\rm{2 Mly}}\)away.

Short answer

The brightness related to the sun is obtained as: \(\eta = {\rm{ 6}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 11\;}}}}\).

Step-by-step solution

Relative brightness.

The relative brightness is given by,

\({\rm{Relative}}\,{\rm{Brightness}} = {\rm{luminosity}} \times \frac{{4\pi r_{sun}^2}}{{4\pi R_{Andromeda}^2}}\)

Here\({r_{sun}}\)is the radius of the sun,\({r_{Andromeda}}\)is the radius of the Andromeda galaxy.

Evaluating the brightness

A star's luminosity is a measurement of the overall amount of light it emits. If we know the star's luminosity, we can use the formula to compute its apparent brightness as:

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

The value of\(L\)is the brightness of the star.

The value of\(d\)is the distance to the star.

Using the formula to evaluate the ratio of apparent brightness for the sun and the Andromeda as:

\(\begin{array}{c}\eta = \frac{{{F_A}}}{{{F_S}}}\\ = \frac{{{L_A}d_S^2}}{{{L_S}d_A^2}}\end{array}\)

Values are then corresponding to Andromeda which is then denoted by the value of\({\rm{A}}\).

The value of\({\rm{S}}\)denoted for the sun.

Putting the values and then we obtain:

\(\begin{array}{c}\eta {\rm{ }} = {\rm{ }}\frac{{{\rm{(1}}{{\rm{0}}^{{\rm{12}}}}{\rm{ \times }}{{\rm{L}}_{\rm{S}}}{\rm{)(1}}{\rm{.58 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ly}}{{\rm{)}}^{\rm{2}}}}}{{{{\rm{L}}_{\rm{S}}}{{{\rm{(2 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ly}} \times {\rm{9}}{\rm{.46}} \times {\rm{1}}{{\rm{0}}^{15}}\,{\rm{m/ly)}}}^{\rm{2}}}}}\\ = {\rm{ 6}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}}\end{array}\)

Therefore, the relative brightness of the Andromeda galaxy with respect to the sun is \({\rm{6}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}}\).