# Estimate the approximate (minimum) mass of an exoplanet in a 3.312 day orbit around a 1.3 solar-mass star, where it is known that the radial velocity caused by orbital motion is 471 metres per second?

May 28, 2017

The planet has about 4 Jupiter masses.

#### Explanation:

${a}^{3} = \frac{G M}{4 {\pi}^{2}} {p}^{2}$

Where $a$ is the semi-major axis, $G$ is the gravitational constant, $M$ is the mass of the star and $p$ is the orbital period.

Now for the Sun $G M = 1.327 \cdot {10}^{20}$, so for a 1.3 solar mass star $G M = 1.725 \cdot {10}^{20}$.

The period is $p = 3.312 \cdot 86400 = 886156.8$ seconds.

Substituting the values gives the semi major $a = 7.099 \cdot {10}^{9}$ metres.

The radial velocity of the planet is ${v}_{p} = \frac{2 \pi a}{p}$.

Rearranging the Kepler equation:

$\frac{4 {\pi}^{2} {a}^{2}}{p} _ 2 = \frac{G M}{a}$

Taking the square root gives:

$\frac{2 \pi a}{p} = {v}_{p} = \sqrt{\frac{G M}{a}}$

Substituting values gives ${v}_{p} = 1.55 \cdot {10}^{5}$ m/s.

The momentum equation relates the sun star mass and velocity to the planet mass and velocity.

$M {v}_{s} = {m}_{p} {v}_{p}$

A solar mass is $1.99 \cdot {10}^{30}$kg. So:

${m}_{p} = \frac{1.3 \cdot 1.99 \cdot {10}^{30} \cdot 471}{1.55 \cdot {10}^{5}} = 7.86 \cdot {10}^{27}$kg.

Jupiter's mass is $1.898 \cdot {10}^{27}$ kg.#

This makes the exoplanet about 4 Jupiter masses.