Chapter 7, Problem 44A

To determine

To Calculate: The mass of the star.

Answer to Problem 44A

  $1.91m_S$

Explanation of Solution

Given:

Orbital period of planet A, $T_A=127.5\text{days}$

Orbital radius of planet A, $r_A=0.615\text{AU}$

Orbital period of planet B, $T_B=520\text{days}$

Orbital radius of planet B, $r_B=1.57\text{AU}$

Formula used:

From Kepler’s third law of planetary motion:

  $T^2=\frac{4{\pi }^2}{Gm}{r}^3$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and m is the mass of the star and T is the orbital period.

Calculations:

From Kepler’s law, one can conclude that the orbital period depends on the mass of the planet and orbital radius.

Considering information given about planet A and planet earth, find the mass of the star.

Orbital period of Earth, $T_E=365.4\text{days}$

Orbital radius of Earth, $r_E=1\text{AU}$

Let mass of Sun be $m_S$

  $\begin{array}{l}{T}_A{}^2=\frac{4{\pi }^2}{Gm}{r}_A{}^3\\ m=\frac{4{\pi }^2}{G{T}_A{}^2}{r}_A{}^3\\ m_S=\frac{4{\pi }^2}{G{T}_E{}^2}{r}_E{}^3\end{array}$

Take the ratio of m and $m_S$ :

  $\begin{array}{l}\frac{m}{m_S}=\frac{r_A{}^3{T}_E{}^2}{T_A{}^2{r}_E{}^3}\\ m=\frac{{\left(0.615\right)}^3{\left(365.4\right)}^2}{{\left(127.5\right)}^2{\left(1\right)}^3}{m}_S\\ m=1.91m_S\end{array}$

Conclusion:

Thus, the mass of the star is 1.91 times the mass of the Sun.