91影视

A system having millions or billions of stars, as well as gas and dust, that are gravitationally bound together.

The connection that can be used to calculate a parent body's mass M from its satellites' orbits:

$\frac{r^3}{T^2}=\frac{G}{4{蟺}^2}M$....................... ( 1 )

If a satellite's r and T are known, the parent's mass M can be estimated, or if r and M are known, T can be determined.

(a)

Putting these values in the equation

$\begin{array}{c}\frac{{\left(5.68脳{10}^{20}\right)}^3}{T^2}=\frac{6.673脳{10}^{-11}}{4{蟺}^2}脳\left(1.52脳{10}^{42}\right)\\ T^2=\frac{6.673脳{10}^{-11}}{4{蟺}^2}脳\frac{\left(1.52脳{10}^{42}\right)}{{\left(5.68脳{10}^{20}\right)}^3}\\ T=\sqrt{\frac{6.673脳{10}^{-11}}{4{蟺}^2}脳\frac{\left(1.52脳{10}^{42}\right)}{{\left(5.68脳{10}^{20}\right)}^3}}\\ T=8.45脳{10}^{15}\text{鈥剆}\end{array}$

Hence, the orbital period of the star is ${\text{8.45脳10}}^{\text{15}}\text{鈥剆}$.

(b)

• The Orbital radius of the star, $r=6.00脳{10}^4\text{鈥凩颈驳丑迟测别补谤s}=6.00脳{10}^4脳9.46脳{10}^{15}\text{鈥刴}=5.68脳{10}^{20}\text{鈥刴}$.
• The orbital period of the star is,$T=6.0脳{10}^7\text{鈥剏}=6.0脳{10}^7\text{鈥剏}脳\frac{365\text{鈥刣}}{1\text{鈥剏}}脳\frac{24\text{鈥删谤}}{1\text{鈥刣}}脳\frac{3600\text{鈥剆}}{1\text{鈥删谤}}=1.89脳{10}^{15}\text{鈥剆}$.

Putting these values in the equation ( 1 ) and we get,

$\begin{array}{c}\frac{{\left(5.68脳{10}^{20}\right)}^3}{{\left(1.89脳{10}^{15}\right)}^2}=\frac{6.673脳{10}^{-11}}{4{蟺}^2}脳M\\ M=\frac{{\left(5.68脳{10}^{20}\right)}^3}{{\left(1.89脳{10}^{15}\right)}^2}脳\frac{4{蟺}^2}{6.673脳{10}^{-11}}\\ M=3.04脳{10}^{43}\text{鈥刱驳}\end{array}$

Hence, the mass of the galaxy for the given orbital period is ${\text{3.04脳10}}^{\text{43}}\text{鈥刱驳}$.