Distance of planets from stars and revolution

In summary, according to Kepler's third law, the planet's orbital period (year) is the same as the earth's.
  • #1
dRyW
7
0
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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  • #2
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.
 
  • #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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
The average orbital radius.
 
  • #7
What he said - actual planetary orbits are ellipses with one focus on the primary so they have a closest and farthest distance.
 
  • #8
Yes, then, what's the average radius for allipses? Is for instance UA the average radius?
 
  • #9
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
 
  • #10
dRyW said:
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.
 
  • #11
@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.
 
  • #12
Simon Bridge said:
@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.
 
  • #13
<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 :)
 
  • #14
Simon Bridge said:
@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.
 
  • #15
@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.
 
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  • #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...
 
  • #17
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".
 
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  • #18
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.

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".

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.
 
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  • #19
Simon Bridge said:
@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. 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.
 
  • #20
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.
 
  • #21
I'll have hard time devicing my answers...

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.

Assume that planet can develop life. Have any edeas about?

More mass the planet have, greater is its gravity. But could be planets bigger then the Earth but with minor or equal mass?
 

1. What is the average distance between planets and stars?

The average distance between planets and stars varies greatly depending on the specific planet and star in question. For example, Mercury, the closest planet to the Sun, has an average distance of about 36 million miles. In contrast, Neptune, the farthest planet from the Sun, has an average distance of about 2.7 billion miles. Overall, the average distance between planets and stars is about 93 million miles, which is the distance between Earth and the Sun.

2. How do scientists measure the distance between planets and stars?

There are several methods that scientists use to measure the distance between planets and stars. One common method is called parallax, which involves measuring the slight change in an object's position relative to background stars as the Earth orbits the Sun. Other methods include using standard candles, which are objects with known luminosity, and using the properties of light to determine distance.

3. How long does it take for a planet to complete one revolution around its star?

The time it takes for a planet to complete one revolution around its star, also known as its orbital period, varies depending on the planet's distance from the star and its orbital speed. For example, Mercury completes one revolution in just 88 days, while Neptune takes 165 years. Earth takes approximately 365 days to complete one revolution around the Sun.

4. Can planets orbit multiple stars?

Yes, it is possible for planets to orbit multiple stars, although it is less common than planets orbiting a single star. These types of systems are known as binary or multi-star systems. An example of a binary star system with orbiting planets is the famous Star Wars planet, Tatooine.

5. How does the distance between a planet and its star affect its revolution?

The distance between a planet and its star plays a crucial role in its revolution. The closer a planet is to its star, the faster it will orbit, due to the stronger gravitational pull from the star. On the other hand, a planet that is farther away from its star will have a slower orbital speed. This relationship between distance and orbital speed is known as Kepler's Third Law of Planetary Motion.

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