If a planet with 2.4 times the mass of Earth was traveling in Earth's orbit, what would its...

Question:

If a planet with 2.4 times the mass of Earth was traveling in Earth's orbit, what would its period be (in years)?

Kepler's Third Law

Kepler's third law states that the orbital period of a planet or a satellite depends upon the orbital radius of the planet or the satellite, Mathematically

{eq}\begin{align} T^2 = \frac{4 \pi^2 r^3}{GM_s} \end{align} {/eq}

Where T is the orbital period, r is the orbital radius and {eq}M_s {/eq} is the mass of the star in the case of the planet or mass of the planet in the case of the satellite.

Answer and Explanation:

Data Given

  • Mass of the planet {eq}m = 2.4 M_E {/eq}
  • Radius of the planet {eq}r = R = 1.5 \times 10^{11} \ \rm m {/eq} where R is the radius of the orbit of the earth.

Now the period of the planet using Kepler's third law

{eq}\begin{align} T^2 = \frac{4 \pi^2 r^3}{GM_s} \end{align} {/eq}

Here the orbital period does not depends upon the mass of the planet so

{eq}\begin{align} T^2 =& \frac{4 \pi^2 R^3}{GM_s} \\ T =& \sqrt{\frac{4 \pi^2 R^3}{GM_s}} \\ T =& \sqrt{\frac{4 \pi^2 (1.5 \times 10^{11} \ \rm m)^3}{6.67 \times 10^{-11} \times 1.98 \times 10^{30} \ \rm kg }} \\ \color{blue}{\boxed{ \ T = T_E = 365 \ \rm days \ }} \end{align} {/eq}

Where {eq}T_E {/eq} is the orbital period of the earth.


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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