To determine the factor by which the mean distance is increased when the orbital period of planet [tex]\( Y \)[/tex] is twice the orbital period of planet [tex]\( X \)[/tex], we start with Kepler’s third law, represented by the equation [tex]\( T^2 = A^3 \)[/tex], where:
- [tex]\( T \)[/tex] is the orbital period of a planet.
- [tex]\( A \)[/tex] is the mean distance from the sun in astronomical units (AU).
Given:
- [tex]\( T_Y = 2 T_X \)[/tex], where [tex]\( T_Y \)[/tex] is the orbital period of planet [tex]\( Y \)[/tex] and [tex]\( T_X \)[/tex] is the orbital period of planet [tex]\( X \)[/tex].
We need to find how much [tex]\( A_Y \)[/tex], the mean distance of planet [tex]\( Y \)[/tex], increases relative to [tex]\( A_X \)[/tex], the mean distance of planet [tex]\( X \)[/tex].
1. Start with the given relationship for planet [tex]\( X \)[/tex]:
[tex]\[
T_X^2 = A_X^3
\][/tex]
2. Given [tex]\( T_Y = 2 T_X \)[/tex], substitute into Kepler’s third law for planet [tex]\( Y \)[/tex]:
[tex]\[
(2 T_X)^2 = A_Y^3
\][/tex]
3. Simplify the left side:
[tex]\[
4 T_X^2 = A_Y^3
\][/tex]
4. From [tex]\( T_X^2 = A_X^3 \)[/tex], substitute [tex]\( A_X^3 \)[/tex] for [tex]\( T_X^2 \)[/tex]:
[tex]\[
4 A_X^3 = A_Y^3
\][/tex]
5. To find the factor by which the mean distance is increased, solve for the ratio [tex]\( \frac{A_Y}{A_X} \)[/tex]:
[tex]\[
\left( \frac{A_Y}{A_X} \right)^3 = 4
\][/tex]
6. Taking the cube root of both sides:
[tex]\[
\frac{A_Y}{A_X} = 4^{1/3}
\][/tex]
Since [tex]\( 4 = 2^2 \)[/tex]:
[tex]\[
4^{1/3} = (2^2)^{1/3} = 2^{2/3}
\][/tex]
Hence, the factor by which the mean distance is increased is:
[tex]\[
2^{2/3}
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{2^{\frac{2}{3}}}
\][/tex]
