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Synodic period closer to 1au

  • #1

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 cant 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!!
 

Answers and Replies

  • #2
gneill
Mentor
20,779
2,759
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?
 
  • #3
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?
 
  • #4
gneill
Mentor
20,779
2,759
Yes.
 

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