Q.Let us assume that our galaxy consists of $2.5 \times 10^{11}$ 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 $10^{5}$ ly.

Concept: Kepler’s Third Law — for a body orbiting a central mass, the square of the orbital period is proportional to the cube of the semi-major axis.

Step 1: Model the galaxy’s mass as concentrated at the centre.

Total mass $M = 2.5 \times 10^{11} M_\odot$, where $M_\odot = 2 \times 10^{30} \, \text{kg}$.

Orbital radius $R = 50{,}000 \, \text{ly}$. Convert to metres:

$1 \, \text{ly} = 9.46 \times 10^{15} \, \text{m}$, so

$R = 5 \times 10^4 \times 9.46 \times 10^{15} = 4.73 \times 10^{20} \, \text{m}$.

Step 2: Apply Kepler’s Third Law in Newtonian form.

For a circular orbit:

$T^2 = \frac{4\pi^2}{G M} R^3$

where $G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$.

Step 3: Compute $T$.

$R^3 = (4.73 \times 10^{20})^3 = 1.058 \times 10^{62} \, \text{m}^3$

$T^2 = \frac{4\pi^2 \times 1.058 \times 10^{62}}{6.67 \times 10^{-11} \times 2.5 \times 10^{11} \times 2 \times 10^{30}}$

Simplify denominator: $6.67 \times 2.5 \times 2 \times 10^{-11+11+30} = 33.35 \times 10^{30} = 3.335 \times 10^{31}$ …

This is a classic problem where you treat the galaxy’s mass as though it’s concentrated at the centre — at least for stars far out in the disk. The key insight is that for a star in a roughly circular orbit around the galactic centre, the gravitational force providing the centripetal acceleration comes from the total mass inside its orbit. Outside mass doesn’t contribute (Newton’s shell theorem). So the star’s motion is exactly like a planet orbiting a point mass.

We can therefore use Kepler’s Third Law in its Newtonian form:

where $T$ is the orbital period, $r$ is the orbital radius, $M$ is the total mass enclosed within that radius, and $G$ is the gravitational constant.

Let’s work through it step by step.

1. Identify the given data.

   • Number of stars inside the orbit: $N = 2.5 \times 10^{11}$ (the problem says “our galaxy consists of …” — we assume this is the total mass, and since the star is at 50,000 ly, which is half the galactic diameter, essentially all the galaxy’s mass lies inside this orbit).
   • Mass per star: one solar mass $M_\odot = 1.99 \times 10^{30}$ kg.
   • Orbital radius: $r = 50,000$ light-years.
   • We’ll need $G = 6.67 \times 10^{-11}$ N·m²/kg².

2. Convert light-years to metres.

   One light-year is the distance light travels in one year:

So

3. Find the total mass $M$ inside the orbit.

4. Apply Kepler’s Third Law.

Plug in numbers:

First compute $GM$:

Then $r^3$:

So

$4\pi^2 \approx 39.48$, so numerator:

Thus

Taking square root:

5. Convert seconds to years. There are $365.25 \times 24 \times 3600 \approx 3.156 \times 10^7$ seconds in a year. …

Step 1: Treat the galaxy's enclosed mass as if concentrated at the centre (valid by the shell theorem for a star at the edge of a roughly symmetric mass distribution) — since the star sits at $r=50{,}000$ ly, exactly the galaxy's radius (diameter $10^5$ ly), essentially the full mass $M=N\times M_\odot$ is enclosed.

Step 2: Compute the enclosed mass: $M=2.5\times10^{11}\times1.99\times10^{30}\approx4.98\times10^{41}$ kg.

Step 3: Convert $r$ from light-years to metres: $1\text{ ly}\approx9.46\times10^{15}$ m, so $r\approx4.73\times10^{20}$ m. …