Chapter 5, Problem 88GP

To determine

To find: The mass of galaxy and number of stars in it.

Answer to Problem 88GP

  $3.45\times {10}^{41}$ kg and $1725\times {10}^8$

Explanation of Solution

Given:

  $\begin{array}{l}r=30,000\text{light years}\\ =30,000\times 9.5\times {10}^{15}\text{m}\end{array}$

  $T=2\times {10}^6$ years

Formula Used:

The Gravitational Attraction provides the centripetal acceleration.

  $\begin{array}{l}\frac{G{M}_{\text{Galavy}}{M}_S}{R^2}=\frac{M_S{V}^2}{R}\\ M_G=\frac{V^{2}R}{G}\end{array}$

But $V=2\pi R/T$

Therefore, $M_G=\frac{4{\pi }^2{R}^3}{T^{2}G}$

Calculation:

Substituting the values

  $M=\frac{4{\pi }^2{\left(30000\times 9.5\times {10}^{15}\right)}^3}{6.67\times {10}^{-11}\times {\left(6.3\times {10}^{15}\right)}^2}$

  $\Rightarrow M=3.45\times {10}^{41}$ kg

and number of stars are

  $N=\frac{3.45\times {10}^{41}}{2\times {10}^{30}}=1725\times {10}^8$

Conclusion:

Therefore the mass of galaxy is $3.45\times {10}^{41}$ kg and number of stars are $1725\times {10}^8$ .