Exoplanet

On the basis of the graph we can read off the period and the absolute value of the speed of the star $\begin{aligned} T &\approx 1.5 \,\text{months} = \frac{1}{8} \,\text{year} \\ v_s &= \sqrt{v_{s,x}^2 + v_{s,y}^2} = \text{max}(v_{s,x}) = 60 \,\text{m}/\text{s} \end{aligned}$ Since the star is similar our sun and has approximately the same mass, we can use the radial distance $a_0 = 1\,\text{AU}$ and orbital period $T_0 = 1\,\text{year}$ of the earth to calculate the semi-axis $a_p$ of the exoplanet according to the third Kepler's law $a_p = a_0 \left( \frac{T}{T_0} \right)^{3/2} = \left( \frac{1}{8} \right)^{3/2} a_0 = \frac{1}{4} \,\text{AU}$ Since the star and the exoplanet circle around a common center of gravity, according to the lever rule, the product of mass and distance must be the same for both: $m a_p = M a_s \quad \Rightarrow \quad \frac{m}{M} = \frac{a_s}{a_p}$ Since the star makes a uniform circular motion, its velocity corresponds to the ratio of circumference and period $v_s = \frac{2 \pi a_s}{T} \quad \Rightarrow \quad a_s = \frac{v_s T}{2 \pi}$ If we combine both equations, the result is $\frac{m}{M} = \frac{a_s}{a_p} = \frac{v_s T}{2 \pi a_p} = \frac{60\,\text{m}/\text{s} \cdot \frac{1}{8} \,\text{year}}{2 \pi \cdot \frac{1}{4} \,\text{AU}} \approx \frac{60 \cdot 3 \cdot 10^{7}}{ 4 \pi \cdot 1.5 \cdot 10^{11}} \approx 10^{-3}$