Given: G = 6.67259 \times 10^{-11} N m^2/kg^2 A planet is orbiting a star. Calculate the mass of...


Given: G = {eq}6.67259 \times 10^{-11} N m^2/kg^2 {/eq} A planet is orbiting a star. Calculate the mass of the star using the fact that the period of the planet is {eq}4.75 \times 10^7 {/eq} s and its distance from the star is {eq}2.95 \times 10^{11} {/eq} m. Answer in units of kg.

Kepler's Third Law:

Kepler's third law tells us how planets orbit stars. As a function of the distance between the celestial bodies r and the mass of the larger body (the one being orbited), we can determine the period of circular motion that these celestial bodies have using the equation:

{eq}\displaystyle T = 2\pi \sqrt{\frac{r^3}{GM}} {/eq}


  • {eq}\displaystyle G = 6.67\ \times\ 10^{-11}\ m^3/kgs^2 {/eq} is the gravitational constant
  • M is the mass of the star being orbited
  • r is the orbital radius

Answer and Explanation:


  • {eq}\displaystyle T = 4.75\ \times\ 10^7\ s {/eq} is the orbital period
  • {eq}\displaystyle r = 2.95\ \times\ 10^{11}\ m {/eq} is the distance between the star and the planet

In order to determine the mass of the star, we will be using:

{eq}\displaystyle T = 2\pi \sqrt{\frac{r^3}{GM}} {/eq}

So here we want M, we square both sides first:

{eq}\displaystyle T^2 = 4\pi^2 \left(\frac{r^2}{GM} \right) {/eq}

We now isolate M by swapping its places with the square of the period:

{eq}\displaystyle M = 4\pi^2 \left(\frac{r^3}{GT^2} \right) {/eq}

So now we substitute:

{eq}\displaystyle M = 4\pi^2 \left(\frac{(2.95\ \times\ 10^{11}\ m)^3}{(6.67\ \times\ 10^{-11}\ Nm^2/kg^2)(4.75\ \times\ 10^7\ s)^2} \right) {/eq}

By evaluating this expression, we thus get:

{eq}\displaystyle \boxed{M = 6.73\ \times\ 10^{30}\ kg} {/eq}

Learn more about this topic:

Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12

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