Question

A planet is in a circular orbit around the sun. It distance from the sun is 4 times the semi major axis of earth's orbits. The period of the planets is$\dots \dots \dots \dots$ earth years

Answer 1

Angular Momentum

• In planetary motion, the planet's net external torque is zero. As a result, angular momentum will not change.

Kepler's third law

T² is proportional to r³, and the orbital period squared is proportional to the distance cubes.

$$\begin{gathered} T^2\alpha r^3 \\ {\left(\frac{T_{earth}}{T_{planet}}\right)}^2={\left(\frac{r_{earth}}{r_{planet}}\right)}^3 \end{gathered}$$

The planet's distance from the sun is 4 times the semi-major axis of the earth's orbit.

$$\begin{gathered} {\left(\frac{T_{earth}}{T_{planet}}\right)}^2={\left(\frac{1}{4}\right)}^3 \\ {T}_{earth}=T_{planet}\times \sqrt{\left(\frac{1}{64}\right)}=T_{planet}\times \left(\frac{1}{8}\right) \\ {T}_{planet}=8\times T_{earth} \end{gathered}$$

$T_{earth}=1year$

$T_{planet}=8years$

Therefore the right answer is 8 years