Answer
See the detailed answer below.
Work Step by Step
a) We know the speed of our solar system (which is 230 km/s) and we know the radius of its circular path around the center of the galaxy (which is 25,000 light-year).
Thus, the period is given by
$$v=\dfrac{2\pi R}{T}$$
$$T=\dfrac{2\pi R}{v}$$
Plugging the known; where $$R=25,000\;\rm ly(\frac{3\times 10^8\;m}{1\;s})(\frac{365.25\;day}{1\;yr})(\frac{24\;h}{1\;day})(\frac{60^2\;s}{1\;h})\\
=\bf2.367\times10^{20}\;\rm m\tag 1$$
$$T=\dfrac{2\pi (2.367\times10^{20})}{230\times 1000}=\bf 6.47\times 10^{15}\;\rm s$$
$$T=\color{red}{\bf2.05\times10^8}\;\rm yr$$
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b)
The number of full rotations is given by
$$N=\dfrac{\Delta t}{T}=\dfrac{5\times 10^{9}}{2.05\times10^8}=24.4\rm \;rev\approx \color{red}{\bf24}\;rev$$
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c)
According to Newton's gravitational law, the force exerted by the center of the Milky Way galaxy on our solar system is given by
$$F=\dfrac{GM_{galaxy} \color{red}{\bf\not}M_{solar}}{R^2}= \color{red}{\bf\not}M_{solar}a_r$$
Hence,
$$ \dfrac{GM_{galaxy} }{R^{ \color{red}{\bf\not}2}} =\dfrac{v^2}{ \color{red}{\bf\not}R}$$
So
$$M_{galaxy}=\dfrac{Rv^2}{G}$$
Plugging the known and plug from (1);
$$M_{galaxy}=\dfrac{(2.367\times10^{20})(230\times 10^3)^2}{6.67\times 10^{-11}}$$
$$M_{galaxy}=\color{red}{\bf 1.88\times10^{41}}\;\rm kg$$
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d) The number of stars is given by
$$N_{star}=\dfrac{M_{galaxy}}{M_{Sun}}=\dfrac{1.88\times10^{41}}{1.99\times 10^{30}}$$
$$N_{star}=\color{red}{\bf 9.43\times 10^{10}}\;\rm Star$$