Let us assume that our galaxy consists of 2.5 x 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky way to be 105 ly.

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To tackle this astronomy problem, we'll need to employ some fundamental concepts. Given that our galaxy consists of approximately 2.5 x 10^11 stars, and we're focusing on the Milky Way, with a diameter of 105 light-years, we can use this information to determine the orbital period of a star located 50,000 light-years from the galactic center.

We can approach this problem using Kepler's third law, which relates the orbital period of a celestial body to its distance from the center of mass it orbits. The law states that the square of the orbital period (T) of a celestial body is directly proportional to the cube of the semi-major axis of its orbit (a).

The semi-major axis of the star's orbit can be considered as the average distance between the star and the galactic center. Given that the star is 50,000 light-years from the galactic center, the semi-major axis (a) would be half of this distance, or 25,000 light-years.

Now, we can use the Milky Way's diameter (105 light-years) to estimate the circumference of its orbit. Since the star is revolving around the center of the Milky Way, its orbit's circumference is essentially the circumference of a circle with a radius of 50,000 light-years. Therefore, the circumference (C) can be calculated using the formula for the circumference of a circle: C = 2πr, where r is the radius.

Using the given values: C = 2 * π * 50,000 ly ≈ 314,159 ly

Now that we have the circumference of the star's orbit, we can find the orbital period (T) using the formula: T^2 ∝ a^3

Since we're comparing two different orbits within the same galaxy, we can use the ratio of their semi-major axes to find the ratio of their orbital periods.

Let's denote T1 as the orbital period of a star at a distance of 50,000 ly and T2 as the orbital period of a star at the edge of the galaxy (at a distance of 52.5 light-years from the center, given the 105 light-year diameter). We can set up the following ratio:

(T1 / T2)^2 = (a1 / a2)^3

Since a1 = 25,000 ly (for the star at 50,000 ly from the center) and a2 = 52.5 ly (for the star at the edge of the galaxy), we can plug these values in:

(T1 / T2)^2 = (25,000 / 52.5)^3

Now, let's solve for T1:

(T1 / T2)^2 = (476.19)^3 T1 / T2 ≈ 53.92

Since T1 / T2 ≈ 53.92, we can assume that the orbital period of the star at a distance of 50,000 ly from the galactic center (T1) is approximately 53.92 times longer than the orbital period of a star at the edge of the galaxy (T2).

Now, let's find the orbital period of a star at the edge of the galaxy. We can use the formula for the orbital period:

T = 2π * r / v

Where: T = Orbital period r = Radius of the orbit v = Orbital speed

For a star at the edge of the galaxy, the radius of the orbit (r) would be 52.5 light-years (half of the galaxy's diameter). The orbital speed (v) can be calculated using the formula:

v = 2π * r / T

Since the star completes one revolution per orbital period (T), we can set up this equation to solve for v. Then, we can use v to find T.

Let's calculate it.