There are a number of options if you want an off-the-shelf solution to fitting RV curves.
Perhaps the best free one is Systemic Console.
However, it is not too hard to do something basic yourself.
First define some terms:
$\nu(t)$ is the true anomaly - the angle between the pericentre and the position of the body around its orbit, measured from the centre of mass focus of the ellipse.
$E(t)$ is the eccentric anomaly and is defined through the equation
$$\tan \frac{E(t)}{2} = \left(\frac{1+e}{1-e}\right)^{-1/2} \tan \frac{\nu(t)}{2}$$
The mean anomaly $M(t)$ is given by
$$M(t) = \frac{2\pi}{p}(t - \tau),$$
where $p$ is the orbital period and $\tau$ is the time of pericentre passage.
"Kepler's equation" tells us that
$$M(t) = E(t) - e \sin E(t)$$
Finally, the radial velocity is given by
$$V_r(t) = K\left[\cos(\omega + \nu(t)) +e \cos \omega \right] + \gamma,$$
where $K$ is the semi-amplitude, $\gamma$ is the centre of mass radial velocity and $\omega$ is the usual angle defining the argument of the pericentre measured from the ascending node.
OK, so the problem is that the radial velocity does not depend explicitly on $t$, but rather on $\nu$. So what you do is the following:
Choose values for $K, \gamma$, $\omega$, $\tau$, $p$ and $e$; these are your "free parameters that describe the orbit. The closer you can get your initial guess, the better.
You use these parameters to predict what the radial velocities would be at the times of observation of your RV datapoints. You do this by calculating $\nu(t)$ using the equations above. Start with the second equation and calculate $M(t)$. Then you have to solve the third equation to get $E(t)$. This is transcendental, so you have to use a Newton-Raphson method or something similar to find the solution. Once you have $E(t)$ then you use the first equation to find $\nu(t)$. Then use 4th equation to calculate $V_r(t)$ at each of your datapoint times.
Calculate a chi-squared (or similar figure of merit) from comparing the predicted and measured values of $V_r(t)$.
Iterate the values of the free parameters and go back to step 2. Continue till your fit converges.
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?
Best Answer
Yes, it's possible to calculate the 3rd if any two or provided. Also, For a given property (like say orbital velocity) Indeed, there are infinite number of values..!
The orbital velocity is calculated by equating the centripetal and gravitational force:
$$\frac{M m}{M+m} \frac{v^2}r=\frac{GMm}{r^2}$$ Where $M$ is the mass of the Sun, $m$ is the mass of the planet, $r$ is the distance between them, and $G$ is the gravitational constant. (The quantity $\frac{M m}{M+m}$ is the reduced (effective) mass.)
For circular orbits, $v=\sqrt{G(M+m)\over r}$ (constant)
Using Newton's generalization of Kepler's third law, the orbital period is given by: $T=2\pi\sqrt{r^3\over G(M+m)}$
See other methods of determining these in olden days...