Physics question - orbit distance to orbit time?

NeilF

NeilF

NeilF

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If a planet is distance D away from its star, and has an orbital period of T. What is the time a planet at 2D and 3D and 4D would take to travel around the star?

Is there a straight relationship or formula even my head would understand?


Given that relationship/formula we should be able to apply it to say Mars (1.52AU, 687 days) or Jupiter (5.2AU, 4332 days) in comparison to Earth's (1AU, 365.2days) distance/period?
 
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I don't think there is a straight up formula for this. The speed of the orbit is going to set the distance from the gravitational centre (the Sun) but many other factors can affect that distance as well (other planets, moons, ascent/descent, etc.).

Or is this hypothetical?
 
Mat said:
The mass of the planets will have a significant effect on orbit time.
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The mass of the orbiting planet makes NO difference. Put a pea in the same orbit around the Sun as Jupiter, and it will take the same time to orbit the Sun as Jupiter.

Remember the hammer and feather on the moon?
 
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SourChipmunk said:
I don't think there is a straight up formula for this. The speed of the orbit is going to set the distance from the gravitational centre (the Sun) but many other factors can affect that distance as well (other planets, moons, ascent/descent, etc.).

Or is this hypothetical?
Click to expand...

There must a simple formula/correlation between distance and orbital period.

If we consider Earth as 1AU (astronomical unit) from the Sun, with an orbital period of 1Y (1 year). Then there must be a simple pattern for the orbital periods at 2AU and 3AU and so on. An exponential graph in effect!?
 
There does not appear to be any pattern because i think there are larger planets that take slower to orbit that are further away and some smaller planets that are closer that are faster to orbit than planets that are larger and further away and the opposite in some cases.
 
SpeedFreak said:
Kepler's second law I believe.
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Thank you! It's his third law. It's the distance cubed and then square rooted.

eg: Earth = 1AU (1 astronomical unit from the sun) and 1Y (takes 1 year to go around it).

So Jupiter at 5.2AU = 5.2 cubed and then square rooted = 11.85 years.
And Saturn at 9.35AU? Well that cubed and then square rooted = 29.31 years.

Those are about right!

Amazing how simple the universe is really :)


groen said:
There does not appear to be any pattern because i think there are larger planets that take slower to orbit that are further away and some smaller planets that are closer that are faster to orbit than planets that are larger and further away and the opposite in some cases.
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See the simple pattern above?

Let's look at Neptune at 30AU. Cube it then square root and you get to 164.3 years. Text books quote 164.7 :) Magic!


ps: And again the size of the planet has nothing to do with its orbit speed/period.
 
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NeilFawcett said:
Thank you! It's his third law. It's the distance cubed and then square rooted.

eg: Earth = 1AU (1 astronomical unit from the sun) and 1Y (takes 1 year to go around it).

So Jupiter at 5.2AU = 5.2 cubed and then square rooted = 11.85 years.
And Saturn at 9.35AU? Well that cubed and then square rooted = 29.31 years.

Those are about right!

Amazing how simple the universe is really :)


See the simple pattern above?

Let's look at Neptune at 30AU. Cube it then square root and you get to 164.3 years. Text books quote 164.7 :) Magic!
Click to expand...
I have learned something today. :)
 
Pluto calculations are the most errant, probably because of it's nasty orbital path. But the rest are really accurate!

orbits.jpg


Of course, every website we visit for the AU and textbook orbit info is probably going to give different numbers. :)
 
Did Mercury earlier! Amazingly accurate, I suspect because least amount of time and most perfect orbit and least other effects.

That calculation gives its orbit around the Sun to within about 6-7 hours accuracy!!
 
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NeilFawcett said:
Thank you! It's his third law. It's the distance cubed and then square rooted.

eg: Earth = 1AU (1 astronomical unit from the sun) and 1Y (takes 1 year to go around it).

So Jupiter at 5.2AU = 5.2 cubed and then square rooted = 11.85 years.
And Saturn at 9.35AU? Well that cubed and then square rooted = 29.31 years.

Those are about right!

Amazing how simple the universe is really :)


See the simple pattern above?

Let's look at Neptune at 30AU. Cube it then square root and you get to 164.3 years. Text books quote 164.7 :) Magic!


ps: And again the size of the planet has nothing to do with its orbit speed/period.
Click to expand...


isnt that only for circular/similar orbits or does it work for different types?


for example are you able to calculate the orbital time of Halley's Comet from earths?
 
Tefal said:
isnt that only for circular/similar orbits or does it work for different types?


for example are you able to calculate the orbital time of Halley's Comet from earths?
Click to expand...

Given my understanding, I'd imagine given a mean orbital distance, it would still work?

EDIT: It's mean distance is around 18.05au, so works out to 76.60yrs... Text book says 75 odd... so not bad!



What I find facinating is it's exactly cubed, not to the power of 3.1 or 3.2, and exactly square rooted etc. Seems soooo perfect!?

And what I find even more fascinating is how microsoft calculator, in scientific mode, has no square root option! (only in basic mode)
 
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^^ Didn't find that covered "our formula"...

This one does from about 9mins in... although it doesn't express it as simply as we have above... ie: In our solar system, cube the AU and square it = orbit in years.

 
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