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In Physics / High School | 2025-07-03

The orbital period, [tex]$P$[/tex], of a planet and the planet's distance from the sun, [tex]$a$[/tex], in astronomical units is related by the formula [tex]$P=a^{\frac{3}{2}}$[/tex]. If Saturn's orbital period is 29.5 years, what is its distance from the sun?

A. 9.5 AU
B. 19.7 AU
C. 44.3 AU
D. 160.2 AU

Asked by Ari08H

Answer (2)

Saturn's distance from the sun can be calculated using the formula P = a 2 3 ​ where P is the orbital period. Given that Saturn's orbital period is 29.5 years, its distance from the sun is approximately 9.5 AU. Therefore, the correct answer is A. 9.5 AU.
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Answered by Anonymous | 2025-07-04

Substitute the given orbital period P = 29.5 into the formula P = a 2 3 ​ .
Raise both sides of the equation to the power of 3 2 ​ to isolate a : a = ( 29.5 ) 3 2 ​ .
Calculate the value: a ≈ 9.547 .
Round to one decimal place to get the final answer: 9.5 AU ​ .

Explanation

Understanding the Problem We are given the formula relating a planet's orbital period, P , and its distance from the sun, a , as P = a 2 3 ​ . We are also given that Saturn's orbital period is P = 29.5 years. Our goal is to find Saturn's distance from the sun, a .

Substituting the Value of P We substitute the given value of P into the formula: 29.5 = a 2 3 ​

Isolating a To solve for a , we raise both sides of the equation to the power of 3 2 ​ : ( 29.5 ) 3 2 ​ = ( a 2 3 ​ ) 3 2 ​

Simplifying the Exponents Simplifying the equation, we get: a = ( 29.5 ) 3 2 ​

Calculating the Distance Calculating the value of ( 29.5 ) 3 2 ​ , we find: a ≈ 9.547

Rounding the Result Rounding the result to one decimal place, we get: a ≈ 9.5 AU

Final Answer Therefore, Saturn's distance from the sun is approximately 9.5 AU.


Examples
Understanding the relationship between a planet's orbital period and its distance from the sun is crucial in astronomy. For example, if we discover a new planet and measure its orbital period, we can estimate its distance from its star using this formula. This helps us understand the layout of planetary systems and the conditions that might exist on other planets. Moreover, this relationship is a direct application of Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. This law is fundamental in understanding celestial mechanics and the dynamics of orbiting bodies.

Answered by GinnyAnswer | 2025-07-04