The Notes: Suppose you observe a star orbiting the galactic center at a speed of 1200 km/s in a circular orbit with a radius of 22 light-days. Calculate the...

Saturday, October 1, 2016

Suppose you observe a star orbiting the galactic center at a speed of 1200 km/s in a circular orbit with a radius of 22 light-days. Calculate the...

Kepler's Third law in Newton's form helps us here. It states that for an object (a star in our case) orbiting another much more massive object (the galactic center)

$\displaystyle\frac{{{P}}^{{2}}}{{{R}}^{{3}}}=\frac{{{4}{\pi}^{{2}}}}{{{G}{M}}},$

where $\displaystyle{P}$ is the orbital period, $\displaystyle{R}$ is the radius of orbit, $\displaystyle{M}$ is the mass of a central object and $\displaystyle{G}\approx{6.7}\cdot{{10}}^{{-{11}}}\frac{{{{m}}^{{3}}}}{{{k}{g{\cdot}}{{s}}^{{2}}}}$ is the gravitational constant.

It is easy to find $\displaystyle{M}$ from this equation, $\displaystyle{M}={{\left(\frac{{{2}\pi}}{{P}}\right)}}^{{2}}\frac{{{R}}^{{3}}}{{G}}.$ We know the speed $\displaystyle{V},$ it is obvious that $\displaystyle{V}=\frac{{{2}\pi{R}}}{{P}},$ so $\displaystyle\frac{{{2}\pi}}{{P}}=\frac{{V}}{{R}}$ and the final formula is

$\displaystyle{M}=\frac{{{{V}}^{{2}}\cdot{R}}}{{G}}.$

To find the numeric answer we have to convert the given radius from light-days into meters. It is $\displaystyle{22}\cdot{2.6}\cdot{{10}}^{{{13}}}{m}$ and the answer is

$\displaystyle\frac{{{{\left({1200000}\right)}}^{{2}}\cdot{22}\cdot{2.6}\cdot{{10}}^{{{13}}}}}{{{6.7}\cdot{{10}}^{{-{11}}}}}\approx{12.3}\cdot{{10}}^{{{36}}}={1.23}\cdot{{10}}^{{{37}}}{\left({k}{g}\right)}.$

The Notes

Saturday, October 1, 2016

Suppose you observe a star orbiting the galactic center at a speed of 1200 km/s in a circular orbit with a radius of 22 light-days. Calculate the...

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