## Distance of planets from stars and revolution

Hi dudes,

My questions are:

What does planet's revolution depends?

If for example is there a planet 4 time the mass of the earth at about its same distance from a star 4 time the mass of sun, could it be it has the same revolution period ot the eath?
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 Recognitions: Homework Help A planet's rotation (day) depends on the total angular momentum of the gas and junk that goes into it's formation. Though some planets get tide-locked to funny ratios of their orbit period which does depend on distance, there is no special reason for the rotation to also be affected and, in general, it isn't. However - it's orbital revolution (year) depends only on the distance from the primary (Kepler's third law) and the mass of the star.
 Sorry, i forgot to specificy that's orbital revolution! Thanks anyway Then according to Kepler's third law in my instance the planet in question would has the same orbital revolution then the earth.

Recognitions:
Homework Help

## Distance of planets from stars and revolution

No - because the mass of the star is also different.

To compute the orbital period you balance the gravitational attraction of the star with the centripetal force of the planet. You can work it out for a circular orbit and substitute the mean radius for an elliptical orbit. It's a good exercise for you to do that yourself.

 You can work it out for a circular orbit and substitute the mean radius for an elliptical orbit
Interesting but i'm an amateur, my brain is easily engulfable, what's the mean radius?
 Recognitions: Homework Help What he said - actual planetary orbits are ellipses with one focus on the primary so they have a closest and farthest distance.
 Yes, then, what's the average radius for allipses? Is for instance UA the average radius?
 Mentor The semi-major axis (or the major axis, which is just twice this value) of the ellipse is important. If you neglect the influence from other planets, the orbital period of a planet depends on the semi-major axis and the reduced mass of the system only. The reduced mass m ]can be calculated from the mass of the planet M1 and the mass of the star M2 via $m=\frac{M_1 M_2}{M_1+M_2}$ - for most systems, it is approximately the mass of the star, and the orbit is independent of the planetary mass. More formulas

 Quote by dRyW Hi dudes, My questions are: What does planet's revolution depends? If for example is there a planet 4 time the mass of the earth at about its same distance from a star 4 time the mass of sun, could it be it has the same revolution period ot the eath?
If the mass of the planet is Mp and the star is M*, then the orbital period scales with

T inversely proportional to √(1+Mp/M*)

which for most star+planet combinations is very nearly the same. However if the planet massed 10 Jupiter masses and the star massed 100, then there's a significant difference.
 Recognitions: Homework Help @qraal: doesn't that imply that the same Mp/M* ration gives the same orbital period regardless of the radius (semi-major axis) of the orbit? I think the question has actually been answered - now I suspect OP wants an actual equation ... @dRyW: have you been asked to derive it perhaps? I gave the classical (approx) version which takes the primary as not moving - you need the reduced mass because bodies actually orbit each other about their common center of mass. At this stage, to answer to your needs, we need to know the context of the question.

 Quote by Simon Bridge @qraal: doesn't that imply that the same Mp/M* ration gives the same orbital period regardless of the radius (semi-major axis) of the orbit? I think the question has actually been answered - now I suspect OP wants an actual equation ...
Hi Simon
The question concerned how mass affected things, not the orbital radius. Notice I said "inversely proportional to" not an equality. I was giving a proportionality not an equation. Hopefully the OP knows the difference.
 Recognitions: Homework Help - question asks generally what the orbit period depends on and follows with an example with included distance as well as masses. Never mind - it is all good stuff :)

Mentor
 Quote by Simon Bridge @qraal: doesn't that imply that the same Mp/M* ration gives the same orbital period regardless of the radius (semi-major axis) of the orbit?
No, larger central masses will give smaller periods. Larger planet masses reduce the period, too, but their effect is much smaller.
 Recognitions: Homework Help @mfb: fine - but not relevant to my question as you have not mentioned radius, which is what I was querying. I am happy that the ratio Mp/M* accounts for the effect of changing either mass correctly. But, by itself, implies that the distances don't matter - which was not what qraal intended. See previous discussion.
 Thanks! Now, soppose there's a planet with plenty time orbital period compared the earth's. For example 20 time, so 20 years. The planet's distance and the dimension of central body must be enough such to let the life exists. How much the planet's average distance or then, the semi-major axis is? Is possible make a simple esteem? Need to have an reasonable standard gravitational parameter? Hope to not forget nothing...
 Recognitions: Homework Help There are too many uncontrolled variables there. You speculate a planet with a 7300day orbit, about a star which is hot enough that this distance is still inside the life zone of the star. Presumably you also want the entire orbit inside the life zone - which limits how eccentric the orbit can be. Comparison of life-zones: http://www.youtube.com/watch?v=j7XJT0N_X2U you need to orbit farther out to get the period longer but then you need a hotter star and they tend to be more massive too so you have to go farther out still. If you are thinking in terms of sci-fi then you should try getting a copy of Steve Jackson Games: GURPS Space - it has a table of life-zones with stars and the formulae needed to compute the orbit period. You can check the orbit periods in the life-zones of each kind... give you a simple rule. Of course not all kinds of stars would have planets "in the wild".