Question

the orbital period, $p$, of a planet and the planet’s distance from the sun, $a$, in astronomical units is related by the formula $p = a^{\frac{3}{2}}$. if saturn’s orbital period is 29.5 years, what is its distance from the sun?
options:
9.5 au
160.2 au
19.7 au
44.3 au

Explanation:

Step1: Start with the given formula

We know the formula relating the orbital period \( P \) and the distance from the sun \( a \) is \( P = a^{\frac{3}{2}} \). We are given that \( P = 29.5 \) years, and we need to find \( a \). So we start with the equation \( 29.5=a^{\frac{3}{2}} \).

Step2: Solve for \( a \)

To solve for \( a \), we can raise both sides to the power of \( \frac{2}{3} \) to isolate \( a \). Using the property of exponents \( (x^m)^n=x^{m\times n} \), we have:
\[
a=(29.5)^{\frac{2}{3}}
\]
First, calculate \( 29.5^{\frac{2}{3}} \). We can rewrite this as \( (\sqrt[3]{29.5})^2 \) or \( (29.5^2)^{\frac{1}{3}} \). Let's calculate \( 29.5^2 = 29.5\times29.5 = 870.25 \). Then we find the cube root of \( 870.25 \). Using a calculator, \( \sqrt[3]{870.25}\approx9.5 \)? Wait, no, wait. Wait, maybe I made a mistake. Wait, let's recalculate. Wait, \( 9.5^3 = 9.5\times9.5\times9.5 = 90.25\times9.5 = 857.375 \), and \( 9.5^{\frac{3}{2}}=\sqrt{9.5^3}=\sqrt{857.375}\approx29.28 \), which is close to 29.5. Wait, maybe the formula is \( P = a^{\frac{3}{2}} \), so if \( P = 29.5 \), then \( a = P^{\frac{2}{3}} \). Let's compute \( 29.5^{\frac{2}{3}} \). Let's take natural logarithm: \( \ln(29.5)\approx3.383 \), multiply by \( \frac{2}{3} \): \( 3.383\times\frac{2}{3}\approx2.255 \), then exponentiate: \( e^{2.255}\approx9.5 \)? Wait, no, wait, maybe the formula is \( P^2=a^3 \) (Kepler's third law, \( P^2 = a^3 \) where \( P \) is in years and \( a \) is in AU). Oh! Wait, maybe I misread the formula. The correct Kepler's third law is \( P^2=a^3 \) (when \( P \) is in years and \( a \) is in AU). So maybe the formula given is \( P = a^{\frac{3}{2}} \), which is equivalent to \( P^2=a^3 \) (since squaring both sides: \( P^2=a^3 \)). So let's use \( P^2 = a^3 \). So \( a^3=P^2 \), so \( a = \sqrt[3]{P^2} \). Given \( P = 29.5 \), so \( P^2 = 29.5^2 = 870.25 \), then \( a=\sqrt[3]{870.25}\approx9.5 \)? Wait, no, \( 9.5^3 = 857.375 \), \( 10^3 = 1000 \), so \( \sqrt[3]{870.25} \) is between 9.5 and 10. Wait, but the options include 9.5 AU, 19.7 AU, 44.3 AU, 160.2 AU. Wait, maybe the formula is \( P = a^{\frac{3}{2}} \), so if \( P = 29.5 \), then \( a = P^{\frac{2}{3}} \). Let's compute \( 29.5^{\frac{2}{3}} \). Let's calculate \( 29.5^{2}=870.25 \), then \( 870.25^{\frac{1}{3}} \). Let's use a calculator: \( \sqrt[3]{870.25}\approx9.5 \) (since \( 9.5^3 = 857.375 \), \( 9.6^3 = 9.6\times9.6\times9.6 = 92.16\times9.6 = 884.736 \), so 870.25 is between 9.5 and 9.6, closer to 9.5). So the answer should be 9.5 AU.

Answer:

9.5 AU