Question

Consider a star moving in a circular orbit at the edge of a galaxy. Construct a problem in which you calculate the mass of that galaxy in kg and in multiples of the solar mass based on the velocity of the star and its distance from the center of the galaxy.

Short answer

The mass of the star is \(3 \times {10^{41}}\,{\rm{kg}}\) and the mass of the star in multiples of solar mass based on the velocity of the star and distance is from center of the galaxy is \(1.5 \times {10^{11}}\).

Step-by-step solution

Newton’s law of gravitation and centripetal force

The Newton’s law of gravitation is given by,

\({F_N} = G\frac{{mM}}{{{r^2}}}\)...... (i)

Here\(m\)and\(M\)are the masses,\(G\)is universal gravitational constant,\(r\)is the distance between them and\(F\)is the force.

The centripetal force is given by,

\({F_G} = m\frac{{{v^2}}}{r}\)....... (ii)

Here\(v\)is the speed.

Evaluating the total mass

On equating (i) and (ii) we get,

\(\begin{align}{F_N} &= {F_G}\\G\frac{{mM}}{{{r^2}}} &= m\frac{{{v^2}}}{r}\\M &= \frac{{{v^2}r}}{G}\end{align}\)

Therefore the expression for the mass of the galaxy is,

\(M = \frac{{{v^2}r}}{G}\)

Substitute\(6.67 \times {10^{ - 11}}\,{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}\)for\(G\),\(2.7 \times {10^5}\,{\rm{m/s}}\)for\(v\)and\(30,000\,{\rm{ly}}\)for\(r\)in the above equation,

\(\begin{align}M &= \frac{{{{\left( {2.7 \times {{10}^5}\,{\rm{m/s}}} \right)}^2}\left( {30,000\,{\rm{ly}}} \right)\left( {9.46 \times {{10}^{15}}\,{\rm{m/ly}}} \right)}}{{6.67 \times {{10}^{ - 11}}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}}}\\ &= 3 \times {10^{41}}\,{\rm{kg}}\end{align}\)

Therefore the mass of the star in multiple of solar mass is,

\(\begin{align}M &= \frac{{3 \times {{10}^{41}}\,{\rm{kg}}}}{{1.99 \times {{10}^{30}}\,{\rm{kg}}}}\\ &= 1.5 \times {10^{11}}\end{align}\)

Therefore the mass of the star is \(3 \times {10^{41}}\,{\rm{kg}}\) and the mass of the star in multiples of solar mass based on the velocity of the star and distance is from center of the galaxy is \(1.5 \times {10^{11}}\).