Answer:
The approximate mass of the Milky Way Galaxy within the Sun's orbit is [tex] 4.73x10^{39}Kg[/tex].
Explanation:
The Universal law of gravitation shows the interaction of gravity between two bodies:
[tex]F = G\frac{Mm}{r^{2}}[/tex] (1)
Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.
For this particular case, M is the mass of the Milky Way Galaxy and m is the mass of the Sun. Since it is a circular motion the centripetal acceleration will be:
[tex]a = \frac{v^{2}}{r}[/tex] (2)
Then Newton's second law ([tex]F = ma[/tex]) will be replaced in equation (1):
[tex]ma = G\frac{Mm}{r^{2}}[/tex]
By replacing (2) in equation (1) it is gotten:
[tex]m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}[/tex] (3)
Therefore, the mass of the Milky Way Galaxy can be determined if M is isolated from equation (3):
[tex]M = \frac{rv^{2}}{G}[/tex] (4)
But r is 27000 light years (r = 27000 ly), Notice that it is necessary to express r in units of meters:
[tex]r = 27000ly \cdot \frac{9.461x10^{15}m}{1ly}[/tex] ⇒ [tex]2.55x10^{20}m[/tex]
Before using equation 4 it is necessary to find the velocity:
[tex]v = \frac{d}{t}[/tex] (5)
Where d is the distance and t is the time, for this case t = 230000000 years.
[tex]t = 230000000years \cdot \frac{31536000s}{1year}[/tex] ⇒ [tex]7.25x10^{15}s[/tex]
Then, the values of d and t can be replaced in equation 5:
[tex]v = \frac{2.55x10^{20}m}{7.25x10^{15}s}[/tex]
[tex]v = 35172m/s[/tex]
Finally, equation 4 can be used:
[tex]M = \frac{rv^{2}}{G}[/tex]
[tex]M = \frac{(2.55x10^{20}m)(35172m/s)^{2}}{6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2}}[/tex]
[tex]M = 4.73x10^{39}Kg[/tex]
Hence, the approximate mass of the Milky Way Galaxy within the Sun's orbit is [tex] 4.73x10^{39}Kg[/tex].
Appendix:
The value of a light year in meters can be determined by means of equation 5:
[tex]v = \frac{d}{t}[/tex]
[tex]d = v.t[/tex]
[tex]d = (3x10^{8}m/s)(31536000s)[/tex]
[tex]d = 9.461x10^{15}m[/tex]
