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What is the average distance from the Sun (in astronomical units) of a planet with an orbital period of 45.66 years?

User Dkong
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Answer:

the average distance from the Sun of a planet with an orbital period of 45.66 years is approximately 12.708 astronomical units (AU).

Step-by-step explanation:

You can calculate the average distance from the Sun (in astronomical units, AU) of a planet using Kepler's Third Law of Planetary Motion, which relates the planet's orbital period (T) to its average distance from the Sun (a). The formula is:

a^3/T^2 = k

Where:

a is the average distance from the Sun in AU.

T is the orbital period in years.

k is a constant.

You need to find "a," so you can rearrange the formula:

a^3 = k * T^2

Now, plug in the values:

a^3 = k * (45.66 years)^2

To solve for "a," take the cube root of both sides:

a = (k * (45.66 years)^2)^(1/3)

You'll need to know the value of the constant "k," which relates to the properties of the Sun and is approximately equal to 1.000041 AU^3/year^2. Now, calculate "a":

a = (1.000041 AU^3/year^2 * (45.66 years)^2)^(1/3)

a ≈ (1.000041 AU^3/year^2 * 2080.9956 years^2)^(1/3)

a ≈ (2081.005 AU^3)^(1/3)

a ≈ 12.708 AU

User Xysun
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