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 the Sun's orbital speed around the center of the Galaxy? How long does it take the Sun to orbit the Milky Way? (In this problem, we assume that the Galaxy can be treated as a single body. Strictly speaking, this isn't correct, but the more elaborate math needed to calculate the problem properly ends up giving almost the same answer.) V = ___ × 10^___ m/s Time = ___ × 10^___ years

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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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