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Distance of planets from stars and revolution

  1. May 27, 2012 #1
    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?
     
  2. jcsd
  3. May 27, 2012 #2

    Simon Bridge

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    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.
     
  4. May 27, 2012 #3
    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.
     
  5. May 27, 2012 #4

    Simon Bridge

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    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.

    Should be about 365/2 days.
     
  6. May 27, 2012 #5
    Interesting but i'm an amateur, my brain is easily engulfable, what's the mean radius?
     
  7. May 27, 2012 #6
    The average orbital radius.
     
  8. May 27, 2012 #7

    Simon Bridge

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    What he said - actual planetary orbits are ellipses with one focus on the primary so they have a closest and farthest distance.
     
  9. May 28, 2012 #8
    Yes, then, what's the average radius for allipses? Is for instance UA the average radius?
     
  10. May 28, 2012 #9

    mfb

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    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 [itex]m=\frac{M_1 M_2}{M_1+M_2}[/itex] - for most systems, it is approximately the mass of the star, and the orbit is independent of the planetary mass.

    More formulas
     
  11. May 28, 2012 #10
    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.
     
  12. May 28, 2012 #11

    Simon Bridge

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    @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.
     
  13. May 29, 2012 #12
    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.
     
  14. May 29, 2012 #13

    Simon Bridge

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    <checks> - 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 :)
     
  15. May 29, 2012 #14

    mfb

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    No, larger central masses will give smaller periods. Larger planet masses reduce the period, too, but their effect is much smaller.
     
  16. May 30, 2012 #15

    Simon Bridge

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    @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.
     
    Last edited: May 30, 2012
  17. May 30, 2012 #16
    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...
     
  18. May 30, 2012 #17

    Simon Bridge

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    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:


    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".
     
    Last edited by a moderator: Sep 25, 2014
  19. May 30, 2012 #18
    Yes, actually my interest for the question is more sci-fi, perhaps i should specify it.

    Yes, i figure it out! A 20 years period needs hotter star.

    Is so impossible esteem the distance? Where can i find this table of line online?

    Aside from this..

    Could a reasonable fraction are outside the life zone but without affect life, or however how life and seasons would affected in these situation? Especially the fraction outside the life zone.

    and let me..what if planet spin much faster then earth? Could this affect (surely of corse) the whole airstreams's development mainly about strength?

    Sorry if it's much sci-fi you not have to reply.
     
    Last edited: May 30, 2012
  20. May 30, 2012 #19

    D H

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    No. What qral meant to say (or should have said) was the orbital period is inversely proportional to √(1+Mp/M*) for a given semi major axis length. The generic formula in Newtonian mechanics for the Keplerian orbital period of some planet p about some star s is
    [tex]\left(\frac{T}{2\pi}\right)^2 = \frac{a^3}{G(M_s+M_p)}[/tex]
    Note that this implies that Kepler's laws are only approximately correct. Kepler's laws as stated by Kepler are good to about 3 decimal places. Newtonian mechanics ups the accuracy considerably, and general relativity improves things even more.
     
  21. May 30, 2012 #20

    mfb

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    The luminosity is rising quicker than the mass, therefore hotter stars may allow an orbital period of 20 years in the habitable zone.

    With some numbers from wikipedia:
    For main-sequence stars, the luminosity L of a star with mass m is approximately [itex]\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^a[/itex] with a~3.5, where the denominators are the sun's values. The radius r of the habitable zone scales with [itex]r \propto \sqrt{L}[/itex] and therefore [itex]r \propto M^{a/2}[/itex] and [itex]T \propto M^{3/2a-1} \approx M^4[/itex], neglecting the mass of the planet. To get an orbital period of 20 years with earth-like conditions, it is enough to have a star with ~2.1 times the stellar mass.
    However, the lifetime of the star is significantly smaller (by a factor of ~5-6), and life would not have 5 billion years to evolve there.
     
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