If a planet existed that had a period twice that of Earth, what would be the length of the semi-major axis

as it orbits the Sun?

4 AU

1.59 AU

2.83 AU

2 AU

1 answer

To determine the semi-major axis of a planet that has an orbital period twice that of Earth, we can use Kepler's Third Law of Planetary Motion, which states that the square of the period of a planet (T) is directly proportional to the cube of the semi-major axis (a) of its orbit. This can be expressed mathematically as:

\[ T^2 \propto a^3 \]

For planets orbiting the same star (in this case, the Sun), we can express this as:

\[ \frac{T^2}{a^3} = k \]

Where \( k \) is a constant.

For Earth:

  • The period \( T \) is 1 year, so \( T_{\text{Earth}} = 1 , \text{year} \).
  • The semi-major axis \( a \) is 1 AU.

For a planet with a period \( T \) that is twice that of Earth:

\[ T_{\text{planet}} = 2 , \text{years} \]

Plugging into Kepler's Third Law:

\[ \frac{(2 , \text{years})^2}{a^3} = \frac{(1 , \text{year})^2}{(1 , \text{AU})^3} \]

Calculating the left side:

\[ \frac{4 , \text{years}^2}{a^3} = \frac{1 , \text{year}^2}{1 , \text{AU}^3} \]

Setting them equal:

\[ \frac{4}{a^3} = 1 \]

This simplifies to:

\[ 4 = a^3 \]

Taking the cube root of both sides:

\[ a = \sqrt[3]{4} \approx 1.59 , \text{AU} \]

Therefore, the semi-major axis of a planet with a period twice that of Earth would be approximately 1.59 AU.