# Astronomical observations of our milky way galaxy indicate that it has a mass about 8.0 *10^{13}...

## Question:

Astronomical observations of our milky way galaxy indicate that it has a mass about 8.0 *10{eq}^{13} {/eq} times the mass of our Sun. A star orbiting on the galaxy edge is about 6.0*10{eq}^4 {/eq} light years a way from the galactic center. What is the orbital period of that star in earth year?

## Objects in Circular Orbits:

When a celestial object is in a circular motion, the gravitational force provides the centripetal force of the motion. When you equate the centripetal force with the gravitational law by Newton, you can describe the motion of the object in terms of the orbital speed, the orbital period, the orbital radius, etc.

## Answer and Explanation:

The orbital period T is given by

{eq}T=\dfrac{2\pi r^{1.5}}{\sqrt{GM}}\\ \,\,\,\, \, \bullet \,r(=6\times 10^4\, Ly) \text{: distance from the the center of the galaxy to the star}\\ \,\,\,\, \, \bullet \,M(=8\times 10^{13} M_s) \text{: mass of the galaxy}\\ \,\,\,\, \, \bullet \,M_s(=1.99\times 10^{30}\, kg) \text{: mass of the sun}\\ \,\,\,\, \, \bullet \, G(=6.67\times10^{-11} N m^2/kg^2)\text{: gravitational constant}\\ {/eq}

Please note, 1 year is {eq}3.16\times 10^7 {/eq} seconds and 1 Ly is {eq}9.46\times 10^{15} {/eq} meters.

So, the orbital period of the galaxy is

{eq}\begin{align} T&=\dfrac{2\pi r^{1.5}}{\sqrt{GM}}\\\\ &=\dfrac{2\pi (6\times 10^4\times 9.46\times 10^{15})^{1.5}}{\sqrt{6.67\times10^{-11}\times 8\times 10^{13}\times 1.99\times 10^{30}}}\\\\ &=8.25\times 10^{14}\, s\\\\ &=\boxed{2.61\times 10^7\, yr} \end{align} {/eq}

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