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?\
\
\\(\bigcirc\\) 9.5 au\
\\(\bigcirc\\) 19.7 au\
\\(\bigcirc\\) 44.3 au\
\\(\bigcirc\\) 160.2 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 solve for \( a \).

Step2: Isolate \( a \)

To solve for \( a \), we can rewrite the formula. First, raise both sides to the power of \( \frac{2}{3} \) to isolate \( a \). So, \( a = P^{\frac{2}{3}} \).

Step3: Substitute \( P = 29.5 \)

Substitute \( P = 29.5 \) into the formula for \( a \): \( a=(29.5)^{\frac{2}{3}} \).

Step4: Calculate the value

First, calculate \( 29.5^{\frac{2}{3}} \). We can do this by first finding the cube root of \( 29.5 \) and then squaring the result, or by squaring \( 29.5 \) and then finding the cube root. Let's use the first method. The cube root of \( 29.5 \) is approximately \( \sqrt[3]{29.5}\approx 3.09 \) (since \( 3^3 = 27 \) and \( 3.1^3=29.791 \), so it's a bit less than \( 3.1 \)). Then square that result: \( (3.09)^2\approx 9.55 \), which is approximately \( 9.5 \) AU. Wait, but let's do it more accurately. Alternatively, use a calculator: \( 29.5^{\frac{2}{3}}=(29.5^{2})^{\frac{1}{3}}=(870.25)^{\frac{1}{3}} \). The cube of \( 9.5 \) is \( 9.5\times9.5\times9.5 = 90.25\times9.5 = 857.375 \), the cube of \( 9.6 \) is \( 9.6\times9.6\times9.6 = 92.16\times9.6 = 884.736 \). Since \( 870.25 \) is between \( 857.375 \) (which is \( 9.5^3 \)) and \( 884.736 \) (which is \( 9.6^3 \)), and closer to \( 9.5^3 \), but wait, maybe I made a mistake in the initial approach. Wait, let's check the options. Wait, maybe I messed up the formula. Wait, Kepler's third law is \( P^2=a^3 \) (when \( P \) is in years and \( a \) is in AU). Oh! Wait, maybe the formula was written incorrectly. Wait, the standard Kepler's third law is \( P^2 = a^3 \), where \( P \) is the orbital period in years and \( a \) is the semi - major axis in AU. So maybe the given formula in the problem was a mis - write, and it should be \( P^2=a^3 \) (which is equivalent to \( P = a^{\frac{3}{2}} \), because if \( P^2=a^3 \), then \( P = a^{\frac{3}{2}} \) when \( P>0 \) and \( a>0 \)). So let's use \( P^2=a^3 \). Then \( a^3=P^2=(29.5)^2 = 870.25 \). Then \( a=\sqrt[3]{870.25} \). Let's calculate \( \sqrt[3]{870.25} \). We know that \( 9.5^3=857.375 \), \( 9.5^3 = 857.375 \), \( 9.51^3=9.51\times9.51\times9.51=(9.51\times9.51)=90.4401\times9.51\approx90.4401\times9 + 90.4401\times0.51=813.9609+46.124451 = 859.085351 \), \( 9.52^3=9.52\times9.52\times9.52=(9.52\times9.52)=90.6304\times9.52\approx90.6304\times9+90.6304\times0.52 = 815.6736+47.127808 = 862.801408 \), \( 9.53^3=9.53\times9.53\times9.53=(9.53\times9.53)=90.8209\times9.53\approx90.8209\times9 + 90.8209\times0.53=817.3881+48.135077 = 865.523177 \), \( 9.54^3=9.54\times9.54\times9.54=(9.54\times9.54)=91.0116\times9.54\approx91.0116\times9+91.0116\times0.54 = 819.1044+49.146264 = 868.250664 \), \( 9.55^3=9.55\times9.55\times9.55=(9.55\times9.55)=91.2025\times9.55\approx91.2025\times9+91.2025\times0.55 = 820.8225+50.161375 = 870.983875 \). Ah, so \( 9.55^3\approx870.98 \), which is very close to \( 870.25 \). So \( \sqrt[3]{870.25}\approx9.55 \), which is approximately \( 9.5 \) AU. Wait, but the options have 9.5 AU as one of the options. Wait, but let's check again. Wait, maybe I made a mistake in the formula. Wait, the original formula is \( P=a^{3/2} \), so if \( P = 29.5 \), then \( a = P^{2/3} \). Let's compute \( 29.5^{2/3} \) using a calculator: \( 29.5^{2/3}\approx(29.5)^{0.6667}\approx e^{0.6667\times\ln(29.5)}\approx e^{0.6667\times3.383}\approx e^{2.255}\approx9.54 \), which is approximately \( 9.5 \) AU. So the answer should be 9.5 AU.

Answer:

9.5 AU