If a planet needs 10 years to orbit the Sun, what is the length of its semi-major axis in AU?

1 Answer
Feb 24, 2016

#\frac{T_E^2}{a_E^3} = \frac{T_x^2}{a_x^3}; \qquad a_x^3 = (\frac{T_x}{T_E})^2a_E^3#
#a_x = (\frac{T_x}{T_E})^{2/3}a_E = (\frac{10}{1})^{2/3}. 1 AU = 4.6416 AU#

Explanation:

Kepler's Second Law : The square of the time period of a planet is proportional to the cube of its semi-major axis distance.
#T^2 \propto a^3 \qquad \rightarrow \qquad \frac{T^2}{a^3} = # constant (same for all planets).

Symbols:
#T_E \equiv 1 Year# - Year is defined as the orbital time period of the Earth around the Sun,
#a_E \equiv 1 AU# - One Astronomical Unit is defined as the average distance between the Earth and the Sun,
#T_x = 10 Years# - Orbital time period of the planet.
#a_x = ?# - The unknown semi-major axis distance.

By Kepler's Second Law,
#\frac{T_E^2}{a_E^3} = \frac{T_x^2}{a_x^3}; \qquad a_x^3 = (\frac{T_x}{T_E})^2a_E^3#
#a_x = (\frac{T_x}{T_E})^{2/3}a_E = (\frac{10}{1})^{2/3}. 1 AU = 4.6416 AU#

Note : Seems to be some object in the Asteroids belt between Mars and Jupiter.