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If we can observe that a star and an unmeasurable planet are in circular orbit around a common center of mass. If we know the speed of the star to be $100\,m/s$, the mass of the star to be $2 \cdot 10^{33}\,g$ and a period of $432,000$ seconds. how can we try to calculate the separation, the mass of the second planet? Can make approximation when appropriate

I'm confused because I'm not sure if there's enough information to solve without making assumptions on the mass or the radius of the second planet. But that's what the question is asking

the only thing I can really calculate here is the orbit radius for the star with the velocity and the period.

Any help would be welcomed.

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closed as off-topic by Rob Jeffries, Kyle Kanos, Jon Custer, Qmechanic Feb 21 '17 at 14:51

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There are different radii that you need to consider (see attached image). The Center Of Mass (COM) is indicated with the red X. The radius of the orbit of the star is $R_s$ and the radius of the orbit of the planet is $R_p$; the total separation between the two is the "semi-major axis" $a = R_p + R_s$.

  1. When you use the velocity and period to find the "radius", which radius is that?

  2. If you use Kepler's law which distance ("radius") does that tell you about?

  3. Based on the definition of the COM (and note that this is also the center of momentum), what other relationships can you make between the planet and the star?

enter image description here

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  • $\begingroup$ When I use the velocity and period to find the "radius", I'm calculating Rs. But I don't quite understand how I can use kepler's third law. I understand that it calculates Rs+Rp, but then we have to make assumptions about planet mass. Could I make the assumption that Mp + Ms = Ms? $\endgroup$ – casualprogrammer Feb 20 '17 at 20:11
  • $\begingroup$ @casualprogrammer you can make that assumption, and it is usually a fair one. I'm not sure if that's what's expected by your instructor. But based on (3) above, you actually don't need to make this approximation. $\endgroup$ – DilithiumMatrix Feb 20 '17 at 21:15
  • $\begingroup$ @casualprogrammer -- You don't have to make any assumptions. Keep it as an unknown value. Kepler's third gives an expression for $(R_s+R_p)^3$ in terms of the known star velocity, period, and mass and the unknown planet mass. The center of mass gives an expression for $(R_s+R_p)$ in terms of the known known star period and mass and the unknown planet mass. That's two equations with two unknowns, $a=R_s+R_p$ and $m_p$. You'll get a cubic. You can make the approximation $M+m\approx M$ at this point to simplify the calculation, but you don't need to (cubics are solvable). $\endgroup$ – David Hammen Feb 20 '17 at 22:55

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