What Happens to a Spacecraft's Synodic Period as Its Orbit Approaches 1 AU?

[wiremonkey22]

Homework Statement

If we were to put a spacecraft into orbit around the Sun at a distance of 0.900 au and then gradually increase the orbital distance closer and closer to 1 au, what would happen to the synodic period of the spacecraft ? Why does this happen?

Homework Equations

1/S = 1/P - 1/E & P2 = a3

The Attempt at a Solution

I get the correct answer when i calculate the known planets, mercury, venus, mars... but my hypothetical planets (orbiting the Sun) are just shy of 1au and just beyond 1au. example planet D (a=0.999) with this formula i get 665.833...(calculating in years) using this 1/s = 1/.9985-1/1. .9985 is what i got with P2 = a3. However when i replace 1/e (earth) with 365.26 and convert that to years it seems reasonable, being 1.001. One explanation has this statement "taking the limit of S as P approaches 1 AU", i just can't seem to find an explanation anywhere on the web. Does this formula break down closer to 1 because we use the Earth as a reference, or am i just using it incorrectly. Please tell me if this is confusing and i will try to explain it better. Thank you!

 

[gneill]

As the orbit of your planet gets closer to that of the Earth, their orbital periods approach one another. That is to say, the difference in speed of the two approaches zero. How long would it take one object to "lap" the other when the differential speed is nearly nil?

 

[wiremonkey22]

Thank you, i think i got it, it helped to understand what synodic period was first instead of just plugging in numbers, once i figured out that the speed would be roughly the same it would take longer for it to return to the same angular position. Am i right here?

 

[gneill]

Yes.

 

[rwisz]

As the spacecraft's orbital distance approaches 1 au, the synodic period will decrease. This is because the synodic period is the time it takes for the spacecraft to return to the same relative position with respect to the Earth and the Sun. When the spacecraft is at 1 au, it is in the same position as the Earth, so the synodic period will be the same as the Earth's orbital period, which is approximately 365.26 days.

The formula you are using, 1/S = 1/P - 1/E, is known as Kepler's Third Law, which relates the orbital period (P) of a planet to its semi-major axis (a). This formula is valid for elliptical orbits, but it may not accurately predict the synodic period when the spacecraft's orbit is very close to the Earth's orbit (1 au). This is because the Earth's orbit is not a perfect circle, so the distance between the Earth and the spacecraft will vary slightly over time, affecting the synodic period.

To accurately calculate the synodic period, you may need to use more complex equations that take into account the eccentricity of the Earth's orbit and the changing distances between the Earth and the spacecraft. You may also need to use numerical methods to solve these equations, as they may not have a simple analytical solution.

Overall, the decrease in the synodic period as the spacecraft's orbit approaches 1 au is due to the Earth's slightly elliptical orbit and the changing distances between the spacecraft and the Earth.