Question

the equation ( t^2 = a^3 ) shows the relationship between a planet’s orbital period, ( t ), and the planet’s mean distance from the sun, ( a ), in astronomical units, au. if planet y is twice the mean distance from the sun as planet x, by what factor is the orbital period increased?
options: ( 2^{\frac{1}{3}} ), ( 2^{\frac{1}{2}} ), ( 2^{\frac{2}{3}} ), ( 2^{\frac{3}{2}} )

Explanation:

Step1: Recall Kepler's Third Law

The given equation is \( T^2 = A^3 \), which is a form of Kepler's Third Law for planets orbiting the Sun (where \( T \) is in years and \( A \) is in AU). Let the mean distance of planet \( X \) be \( A_X \) and its orbital period be \( T_X \). For planet \( Y \), the mean distance \( A_Y = 2A_X \), and its orbital period is \( T_Y \).

Step2: Apply the Law to Both Planets

For planet \( X \): \( T_X^2 = A_X^3 \)
For planet \( Y \): \( T_Y^2 = A_Y^3=(2A_X)^3 = 8A_X^3 \)

Step3: Find the Ratio of Periods

We can express \( T_Y^2 \) in terms of \( T_X^2 \). From the equation for \( X \), \( A_X^3=T_X^2 \). Substitute this into the equation for \( Y \):
\( T_Y^2 = 8T_X^2 \)
Take the square root of both sides: \( T_Y = T_X\sqrt{8}=T_X\cdot2\sqrt{2} \)? Wait, no, wait. Wait, the equation is \( T^2 = A^3 \), so \( T = A^{\frac{3}{2}} \). So if \( A_Y = 2A_X \), then \( T_Y=(2A_X)^{\frac{3}{2}}=2^{\frac{3}{2}}A_X^{\frac{3}{2}} \). But \( T_X = A_X^{\frac{3}{2}} \), so the factor is \( 2^{\frac{3}{2}} \)? Wait, no, the options are \( 2^{\frac{1}{3}}, 2^{\frac{1}{2}}, 2^{\frac{2}{3}}, 2^{\frac{3}{2}} \). Wait, let's re - express the ratio.

From \( T^2 = A^3 \), we can solve for \( T \): \( T = A^{\frac{3}{2}} \). Let \( A_X = A \), then \( A_Y = 2A \).

\( T_X = A^{\frac{3}{2}} \)
\( T_Y=(2A)^{\frac{3}{2}}=2^{\frac{3}{2}}A^{\frac{3}{2}} \)

The factor by which the period increases is \( \frac{T_Y}{T_X}=\frac{2^{\frac{3}{2}}A^{\frac{3}{2}}}{A^{\frac{3}{2}}}=2^{\frac{3}{2}} \)? Wait, no, wait the options are \( 2^{\frac{1}{3}},2^{\frac{1}{2}},2^{\frac{2}{3}},2^{\frac{3}{2}} \). Wait, maybe I made a mistake. Wait, the equation is \( T^2 = A^3 \), so if we let \( A_Y = 2A_X \), then:

\( T_Y^2=(2A_X)^3 = 8A_X^3 \)
But \( T_X^2 = A_X^3 \), so \( T_Y^2 = 8T_X^2 \)
Then \( T_Y=T_X\sqrt{8}=T_X\cdot2\sqrt{2} \)? No, \( \sqrt{8}=2\sqrt{2}=2^{1 + \frac{1}{2}}=2^{\frac{3}{2}} \). Yes, because \( \sqrt{8}=\sqrt{2^3}=2^{\frac{3}{2}} \). So the factor is \( 2^{\frac{3}{2}} \).

Answer:

\( 2^{\frac{3}{2}} \) (the last option among the given options, which is \( 2^{\frac{3}{2}} \))