Monday 9 June 2014

Assuming that the mass of the Milky Way Galaxy is `10^11` times that of the Sun and that the Sun is `2.45 × 10^20` meters from its center, what is...

Hello!


Kepler's first law states that an orbit of a planet around the Sun is an ellipse with the Sun being at one focus of it. Our problem is close enough: the Milky Way Galaxy may be considered as a sun (large central body), and the Sun may be considered as a planet orbiting it. I assume this ellipse is almost a circle, because in the opposite case the speed of the Sun would be different...

Hello!


Kepler's first law states that an orbit of a planet around the Sun is an ellipse with the Sun being at one focus of it. Our problem is close enough: the Milky Way Galaxy may be considered as a sun (large central body), and the Sun may be considered as a planet orbiting it. I assume this ellipse is almost a circle, because in the opposite case the speed of the Sun would be different at the different points of its trajectory.


Denote the radius of this orbit as `R,` it is given, and the speed as `V,` it is unknown. Also denote the given mass of the Milky Way Galaxy as `M` and the Sun's mass as `m,` it is about `2*10^(30) kg.` Then the Newton's Law of Universal Gravitation tells us that the gravity force `F` is `G*(M*m)/R^2,` so by Newton's second law the acceleration is `F/m = (GM)/R^2.`


We also know that the acceleration of a body in uniform circle motion is `V^2/R,` so `(GM)/R^2 = V^2/R,` or `V^2 = (GM)/R,` or `V = sqrt((GM)/R) = sqrt((G*10^11*m)/R).`


Numerically it is about


`sqrt((6.7*10^(-11)*10^11*2*10^30)/(2.45*10^20)) = sqrt(5.5*10^10) approx 2.3*10^5 (m/s),`


or about 230 km/s.


The time needed is about `(2piR)/V approx 6.7*10^15 (s) approx` 212,000,000 years.


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