Consider a planet and a sun rotating about their center of mass. The sun has a mass 0.9 of the mass of our sun.

I must calculate the planet's mass in terms of jupiter's mass and the planets orbital radius in terms of earth's radius – I don't understand what this radius means. Aren't orbits elliptical, or are orbits about a center of mass actually circular?

I know that the period of orbit of the sun in the problem is 1500 days and its maximum line of sight velocity is $70m/s$ and minimum is $-70m/s$

I've tried a bunch of approaches but can't get anything to work.

First, the mass and velocity of the planet are $m_p, v_p$ and of the star $m_s, v_s$. Then conserving angular momentum $$L_{s_i}+L_{si} = L_{pf} + L_{sf}$$ It is also true that the total mass of the system $M = m_p + m_s$. Additionally, we can describe the period of orbit with the equation

$T^2GM = 4\pi^2r^3$ where $r$ is the radius of orbit.

Now for the first equation, we don't actually know what the final velocity of the star is unless we introduce kinetic and potential energy.

Then writing out the (very long) equation for energy

$\frac{1}{2}m_pv_{pi}^2 + \frac{1}{2}m_sv_{si}^2 + U_{pi} + U_{si} = \frac{1}{2}m_pv_{pf}^2 + \frac{1}{2}m_sv_{sf}^2 + U_{pf} + U_{sf}$

We want to solve for the mass and radius of rotation of the planet. We have four unknowns $v_{pi,f}$, $r_p, r_s$ and have three equations. What do I do? Edit

The fourth equation can be written down as $a_{red} = \frac{v^2}{r}$ where r is the reduced radius and the acceleration is that of the reduced mass.

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  • $\begingroup$ Want to check a couple of points. First, you are assuming a circular orbit, (otherwise, "radius" of the orbit makes no sense). Also when you say "The sun has a mass 0.9 of the mass of the sun" you mean "the sun in this problem... the sun in Earth's solar system", right? Finally for this: "I know that the period of orbit of the sun is 1500 days" - do you mean "orbit of this planet, and this sun"? $\endgroup$
    – Floris
    Commented Dec 15, 2016 at 21:22
  • $\begingroup$ If you can ignore the fact that you are orbiting about the barycenter, you can ignore the mass. But when your planet has a large mass, it affects the location of the apparent center of rotation (vs the distance to the sun). Could this be what gives you the missing equation? $\endgroup$
    – Floris
    Commented Dec 15, 2016 at 21:23
  • $\begingroup$ Finally - what do you call the "line of sight velocity"? Is that $\omega r'$ where $r'$ is the reduced radius (to the barycenter)? $\endgroup$
    – Floris
    Commented Dec 15, 2016 at 21:24
  • $\begingroup$ Yes the line of sight velocity is the reduced radius $r'$. Regarding your second comment, would we then write the equation $x_{com} =\frac{d_pm_p+d_sm_s}{M}$ where $d_p$ and $d_s$ are the distances to the center of the planet and the sun? $\endgroup$ Commented Dec 15, 2016 at 21:30
  • $\begingroup$ See if this article on reduced mass helps... $\endgroup$
    – Floris
    Commented Dec 15, 2016 at 21:31

1 Answer 1


I don't understand what this radius means. Aren't orbits elliptical, or are orbits about a center of mass actually circular?

I think that you will have to assume a circular orbit - otherwise, "radius" has no meaning as you correctly point out.

As for the main body of your question - I think you are making this harder than it needs to be.

You are asked to compute the radius in terms of Earth's radius. Now we know from Kepler's Laws that

$$T^2 GM = 4 \pi^2 r^3$$

According to this article on gravitational two-body problems, for the case where the mass of the planet is not negligible compared to the mass of the "sun", we simply replace $M$ with $M+m$; now we can write down the first equation for the mass and orbital radius of the planet:

$$\frac{T_e^2 G M_s}{r_e^3} = \frac{T_p^2~ G ~(0.9~ M_s + m_p)}{r_p^3}\tag1$$

There are only two unknowns in this equation: $r_p$ and $m_p$.

Next, since you have the line of sight velocity and the period, the distance $r_p$ of the planet to the center of rotation follows immediately:

$$\begin{align}v &= \omega r_p = \frac{2\pi r_p}{T}\\ \Rightarrow r_p &= \frac{vT}{2\pi}\end{align}\tag2$$

Now that you have $r_p$ you can substitute into (1) and that should give you your answer.

Converting these to the radius of Earth and the mass of Jupiter should be straightforward. It might be worth calculating how different the answer would be if you could assume the planet to be "light" - then the simple case of Kepler's law would tell us

$$\frac{r_p}{r_e}=\left(\frac{1500}{365.25}\right)^{\frac23}\approx 2.56$$

This is not the answer to your question - but how far off is it from the answer you get from the above? Before you do the calculation askyourself this - do you expect the number to be bigger or smaller?


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