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How is orbital period calculated if perihelion and aphelion are known? For example, the orbit of a spacecraft about the sun has a perihelion distance of 0.5 AU and an aphelion of 3.5 AU., what is its orbital period?

Answer 1

Using Kepler's 3rd law, we get an orbital period of about #2.83# years.

Kepler's third law, which states that the square of the orbital period is proportional to the cube of the average distance from the sun, can be used to calculate the orbital period.

#T^2 prop R^3#

Where #T# is the orbital period and #R# is the body's average distance from the sun, or the semi-major axis. Furthermore, this ratio is the same for any body that orbits the sun. We can therefore rewrite this expression as a ratio in terms of some constant, #C#.

#T^2 / R^3 = C#

We know that the orbital period for the Earth is #1 " year"#, and the Earth's semi-major axis is defined as #1 " AU"#. Using this information we can solve for #C#.

#C = T_"Earth"^2/R_"Earth"^3 = (1 " year")^2/(1 " AU")^3 = 1 "year"^2/"AU"^3#

Remember that #C# is the same for all bodies orbiting the Sun. We can calculate the semi-major axis for the spacecraft by taking the average of its perihelion and aphelion.

#R_"spacecraft" = ("perihelion" + "aphelion")/2 #

#R_"spacecraft" = (.5 " AU" + 3.5 " AU")/2 #

#R_"spacecraft" = 2 " AU"#

We can now determine the orbital period.

#T^2 = CR^3 #

#T^2 = (1 "year"^2/"AU"^3)(2 " AU")^3#

#T^2 = 8 " years"^2#

#T = 2 sqrt(2) " years"#

So the spacecraft has an orbital period of about #2.83# years.

*Note: Kepler's 3rd law works for things orbiting bodies other than the sun, but the constant, #C# will be different.

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Answer 2

Levi Anderson

The orbital period (P) of a spacecraft around the Sun can be calculated using Kepler's third law of planetary motion:

P^2 = a^3

Where:

P is the orbital period in years
a is the semi-major axis of the orbit in astronomical units (AU)

Given:

Perihelion distance (r_p) = 0.5 AU
Aphelion distance (r_a) = 3.5 AU

The semi-major axis (a) can be calculated using the formula: a = (r_p + r_a) / 2

Substituting the given values: a = (0.5 AU + 3.5 AU) / 2 = 4 AU / 2 = 2 AU

Now, substituting the value of a into Kepler's third law: P^2 = 2^3 P^2 = 8

Taking the square root of both sides: P = √8 ≈ 2.83 years

Therefore, the orbital period of the spacecraft around the Sun is approximately 2.83 years.

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