A planet of mass m = 4.55*10^{24} kg is orbiting in a circular path a star of mas M =...


A planet of mass m = 4.55*10{eq}^{24} {/eq} kg is orbiting in a circular path a star of mas M = 1.75*10{eq}^{29} {/eq} kg. The radius of the orbit is r = 8.55*10{eq}^7 {/eq} km. What is the orbital period (in Earth days) of the planet Tplanet?

Circular Orbit of the Planet :

When a planet revolves around the star in the circular orbit, the gravitational force acting between the planet and the star acts as the centripetal force. The expression of the relation between the time period and the radius of the orbit is given by as following ;

{eq}T^2 = \dfrac{4\ \pi^2}{G\ M} \ r^3 {/eq}


  • T is the time period.
  • G is the gravitational constant.
  • M is the mass of the star.
  • r is the radius of the orbit.

Answer and Explanation:

Given :

  • Mass of the star {eq}M = 1.75 \times10^{29} kg {/eq}
  • Radius of the orbit {eq}\ r = 8.55 \times 10^7\text{ km } = 8.55 \times 10 ^{10}\ m {/eq}
  • Gravitational constant {eq}\ G = 6.67\times 10^{-11}N.m^2/kg^2 {/eq}

From Kepler's third law equation ;

{eq}T^2 = \dfrac{4\ \pi^2}{G\ M} \ r^3 {/eq}

Putting values in the above equation, we get;

{eq}T^2 = \dfrac{4 \pi ^2}{(6.67\times 10^{-11}N.m^2/kg^2)\times( 1.75 \times10^{29} kg)} \times (8.55 \times 10 ^{10}\ m)^3\\ T^2 = 2.11 \times 10^{15}\ s^2 {/eq}


{eq}T =45977684.8 \ s {/eq}


24 hour = 1 Earth's day


{eq}24 \times 3600 \ s = 1\ \text{ Earth's day } {/eq}


{eq}1 s = \dfrac{1}{24 \times 3600 \ s }1\ \text{ Earth's day } {/eq}

Then in our case ;

{eq}45977684.8 \ s = 45977684.8 \ s \times \dfrac{1}{24 \times 3600 \ s } \times1\ \text{Earth's day } = 532.1 \ \text{Earth's day } {/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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