# 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.

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.