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alexandr402 [8]
3 years ago
13

A planet requires 1200(Earth) days to complete it's circular orbit around it's hosting star. The planet's orbital radius is abou

t 3 AU. What is the mass of the hosting star?
Physics
1 answer:
12345 [234]3 years ago
7 0
Let's call M the mass of the star and m the mass of the planet, which should cancel.  The gravitational force should equal the centripetal force:

\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}

M=rv^2/G

We compute v = \dfrac{2 \pi r}{T}

M=\dfrac{4 \pi^2 r^3}{G T^2}  

This is Keppler's Law.  We convert to MKS as we plug in and get

M=\dfrac{4\pi^2(3 \cdot 1.496 \times 10^{11}) ^3}{6.674\times 10^{-11} ( 1200 \cdot 24 \cdot 60 \cdot 60 )^2 }

M=4.9744\times 10^{30} kg

which is about two and a half solar masses if I haven't made any errors.




  
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