Two equal-mass stars maintain a constant distance apart of \({\bf{8}}{\bf{.0}} \times {\bf{1}}{{\bf{0}}^{{\bf{11}}}}\;{\bf{m}}\) and revolve about a point midway between them at a rate of one revolution every 12.6 yr. (a) Why don’t the two stars crash into one another due to the gravitational force between them? (b) What must be the mass of each star?

The star'mass can be calculated by equating the gravity force with the centripetal force.

The distance between the stars is \(d = 8.0 \times {10^{11}}\;{\rm{m}}\).

The time taken in revolution is \(T = 12.6\;yr\).

(a)

The figure below represents the free body diagram, with representation of force.

Due to circular motion, the two stars are not going to crash. The centripetal force is acting on both the stars, which is maintaining the circular motion.

Another point of consideration is velocity. The velocity of star is changing due to gravity force. In the circular motion if the stars come to rest, then there is possibility of carsh.

(b)

The equation for velocity in terms time period is:

\(v = \frac{{2\pi r}}{T}\)

The gravity force on one star must be equal to the centripetal force is given as,

\(\begin{aligned}G\frac{{\left( M \right)\left( M \right)}}{{{d^2}}} &= M\left( {{v^2}} \right)\left( {\frac{1}{{\frac{d}{2}}}} \right)\\\frac{{G{M^2}}}{{{d^2}}} &= M{\left( {\frac{{2\pi r}}{T}} \right)^2}\left( {\frac{2}{d}} \right)\\\frac{{G{M^2}}}{{{d^2}}} &= \frac{{2M}}{d}{\left( {\frac{{\pi d}}{T}} \right)^2}\\M &= \frac{{2{\pi ^2}{d^3}}}{{G{T^2}}}\end{aligned}\)

Substitute values in the above equation,

\(\begin{aligned}M &= \frac{{2{\pi ^2}{{\left( {8 \times {{10}^{11}}\;{\rm{m}}} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right){{\left( {12.6\;{\rm{yr}} \times \frac{{3.15 \times {{10}^7}}}{{1\;{\rm{yr}}}}{\rm{s}}} \right)}^2}}}\\ &= 9.6 \times {10^{29}}\;{\rm{kg}}\end{aligned}\)

Thus, the mass of the star is \(9.6 \times {10^{29}}\;{\rm{kg}}\).