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 the orbital period of planet y is twice the orbital period of planet x, by what factor is the mean distance increased? (options are partially visible with circles above them)

Explanation:

Step1: Define Variables for Planets

Let \( T_X \) and \( A_X \) be the orbital period and mean distance for planet X, and \( T_Y \) and \( A_Y \) be those for planet Y. Given \( T_Y = 2T_X \).

Step2: Apply the Kepler's Law Formula

The formula is \( T^2 = A^3 \). For planet X: \( T_X^2 = A_X^3 \). For planet Y: \( T_Y^2 = A_Y^3 \).

Step3: Substitute \( T_Y = 2T_X \)

Substitute into Y's equation: \( (2T_X)^2 = A_Y^3 \), which simplifies to \( 4T_X^2 = A_Y^3 \).

Step4: Replace \( T_X^2 \) with \( A_X^3 \)

From X's equation, \( T_X^2 = A_X^3 \), so substitute into Y's equation: \( 4A_X^3 = A_Y^3 \).

Step5: Solve for \( \frac{A_Y}{A_X} \)

Take the cube - root of both sides: \( A_Y = A_X \times \sqrt[3]{4} \approx A_X\times1.5874 \). Wait, no, let's do it properly. From \( T^2 = A^3 \), we can express \( A = T^{\frac{2}{3}} \). So for X, \( A_X = T_X^{\frac{2}{3}} \), for Y, \( A_Y = T_Y^{\frac{2}{3}}=(2T_X)^{\frac{2}{3}} = 2^{\frac{2}{3}}T_X^{\frac{2}{3}}=2^{\frac{2}{3}}A_X \). Since \( 2^{\frac{2}{3}}=\sqrt[3]{4}\approx1.587 \), but wait, maybe I made a mistake. Wait, the formula is \( T^2 = A^3 \), so \( A = T^{\frac{2}{3}} \). If \( T_Y = 2T_X \), then \( A_Y=(2T_X)^{\frac{2}{3}} = 2^{\frac{2}{3}}T_X^{\frac{2}{3}}=2^{\frac{2}{3}}A_X \). But \( 2^{\frac{2}{3}}=\sqrt[3]{4}\approx1.587 \). But let's check with the formula \( T^2 = A^3 \). Let's assume \( T_X = 1 \), then \( A_X = 1 \) (since \( 1^2=1^3 \)). Then \( T_Y = 2 \), so \( A_Y^3=T_Y^2 = 4 \), so \( A_Y=\sqrt[3]{4}\approx1.587 \). But the options (from the blurry image, but common question) - wait, maybe I misread the formula. Wait, Kepler's third law is \( T^2\propto A^3 \), so \( \frac{T_1^2}{T_2^2}=\frac{A_1^3}{A_2^3} \). So \( \frac{T_Y^2}{T_X^2}=\frac{A_Y^3}{A_X^3} \). Given \( T_Y = 2T_X \), so \( \frac{(2T_X)^2}{T_X^2}=\frac{A_Y^3}{A_X^3} \), \( 4=\frac{A_Y^3}{A_X^3} \), so \( \frac{A_Y^3}{A_X^3}=4 \), then \( \frac{A_Y}{A_X}=\sqrt[3]{4}\approx1.587 \), but if the options are like \( 2^{\frac{2}{3}} \) or \( \sqrt[3]{4} \), but maybe the formula was mis - remembered. Wait, no, the correct Kepler's third law is \( \frac{T^2}{A^3}=constant \), so \( T_1^2A_2^3=T_2^2A_1^3 \). So \( \frac{A_2^3}{A_1^3}=\frac{T_2^2}{T_1^2} \), \( (\frac{A_2}{A_1})^3 = (\frac{T_2}{T_1})^2 \), \( \frac{A_2}{A_1}=(\frac{T_2}{T_1})^{\frac{2}{3}} \). Since \( T_2 = 2T_1 \), \( \frac{A_2}{A_1}=2^{\frac{2}{3}}=\sqrt[3]{4}\approx1.587 \), but if the options are \( \sqrt[3]{4} \) or \( 2^{\frac{2}{3}} \), but maybe the original problem has a typo or I misread the formula. Wait, maybe the formula is \( T = A^3 \)? No, the standard is \( T^2=A^3 \) (when T is in years and A is in AU for solar system). So according to this, the factor by which the mean distance of Y is greater than X is \( \sqrt[3]{4}\approx1.587 \), which is \( 2^{\frac{2}{3}} \) or \( \sqrt[3]{4} \). But if we consider the formula \( T^2 = A^3 \), and we want to find the ratio \( \frac{A_Y}{A_X} \) when \( T_Y = 2T_X \), we have:

From \( T_X^2=A_X^3 \) and \( T_Y^2 = A_Y^3 \), divide the second equation by the first: \( \frac{T_Y^2}{T_X^2}=\frac{A_Y^3}{A_X^3} \)

Substitute \( T_Y = 2T_X \): \( \frac{(2T_X)^2}{T_X^2}=\frac{A_Y^3}{A_X^3} \)

Simplify left - hand side: \( 4=\frac{A_Y^3}{A_X^3} \)

Take the cube root of both sides: \( \frac{A_Y}{A_X}=\sqrt[3]{4}\approx1.587 \)

But if we made a mistake in the formula, and the formula was \( T = A^3 \) (which is wrong, but just for checking), then \( A=\sqrt[3]{T} \), and \( A_Y=\sqrt[3]{2T_X}=\sqrt[3]{2}A_X\approx1.26A_X \), but that's not the case.

Wait, maybe the original problem has the formula \( T^2 = A^3 \), and the answer is \( \sqrt[3]{4} \) or \( 2^{\frac{2}{3}} \). But let's re - do the calculation:

Let \( T_X = t \), so \( T_Y = 2t \)

For planet X: \( t^2=A_X^3\Rightarrow A_X = t^{\frac{2}{3}} \)

For planet Y: \( (2t)^2=A_Y^3\Rightarrow4t^2 = A_Y^3\Rightarrow A_Y=(4t^2)^{\frac{1}{3}}=4^{\frac{1}{3}}t^{\frac{2}{3}} \)

Since \( A_X = t^{\frac{2}{3}} \), then \( A_Y = 4^{\frac{1}{3}}A_X=\sqrt[3]{4}A_X\approx1.587A_X \)

So the factor is \( \sqrt[3]{4} \) (or \( 2^{\frac{2}{3}} \))

Answer:

The factor by which the mean distance of planet Y is greater than that of planet X is \( \sqrt[3]{4} \) (or approximately \( 1.59 \))