# Synodic period closer to 1au

• wiremonkey22
In summary, the synodic period of a spacecraft orbiting the Sun would decrease as the orbital distance is gradually increased closer to 1 AU. This is due to the orbital periods of the spacecraft and Earth approaching each other as their orbits get closer. As the difference in speed between the two objects decreases, it takes longer for one to "lap" the other, resulting in a longer synodic period. This can be seen in the equation 1/s = 1/P - 1/E, where the orbital period of Earth (E) is used as a reference.

## 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!

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?

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?

Yes.