What main factors determine the orbital period of planets in a solar system? Do the planet’s size and distance from the sun have anything to do with it?
Would it be possible for the Earth to orbit the sun twice as fast, or half as fast, and still maintain the same orbital path? Likewise, if the Earth suddenly doubled (or halved) in mass, would its orbital path and/or the length of the sidereal year be affected at all?
It’s determined solely by the mass of the star (sun) at the centre, and the distance between the star and the planet. If the Earth were twice as large, or half as large, and were still travelling at the same speed, the orbit would be the same.
It’s all about the distance to the star. In the right units, the relationship is T[sup]2[/sup] = a[sup]3[/sup], where T is the period (in years) and a is the semimajor axis of the eliiptical orbit (that’s one half the widest diameter of the ellipse), in astronomical units.
So for instance, Pluto is about 40 AU from the Sun. 40[sup]3[/sup] is 64000. Square root of 64000 is about 250, which is roughly the period of Pluto in years. I’m rounding off all over the place here, but the relation works.
You can use other units (miles, minutes, etc.) but then you have to introduce a conversion factor.
So if we were to give the Earth a “push” to make it travel faster, then it would eventually slow back down into its normal orbit?
Doesn’t that assume the star has the same mass as the Sun? The OP talked about “a solar system”, not “the Solar System”, so the sun need not be the Sun.
No, its orbit would become more elliptical: intially it would move further away from the Sun, then it would come back in closer to the Sun. This would repeat itself indefinitely,
On the other hand, if the Earth were travelling in a similarly shaped orbit, but closer to the Sun, then it would be travelling faster.
Nope, that’s one of Kepler’s Third Law of Planetary Motion, and it’s universal.
So, if the Solar System was replaced by two bodies – a “sun” with the mass of the Earth, and a planet also with the mass of the Earth, separated by 1 AU, each would be in orbit around the other with a period of a year? That’s hard to believe, even without doing the sums – if they were in circular orbits, it would be around the centre of mass, and because the mass is som much less, the oribital period would be many times longer.
Well, I’m now getting into the area where I’m having to look up stuff to be sure I know what I’m talking about. Astrophysics 101 was a long time ago. 
Kepler’s laws predate Newton (but can be derived from Newton’s Laws), and are as such a bit simplified. As long as the orbiting body’s mass is negligable compared to the star, and as long as there aren’t any other big gravitational influences around, the relationship holds.
The proportionality a[sup]3[/sup] = k T[sup]2[/sup]always holds, yes, but the proportionality constant k does in fact depend on the mass of the central star. Specifically, k = G M / 4 pi[sup]2[/sup], where G is Newton’s constant and M is the sum of the two masses (which is, for all intents and purposes, equivalent to the mass of the central body.)
Ah, thanks MikeS. I was afraid I was forgetting something, and I was!
Well, now we get into definitions. The period would still be a “year”, by definition, but it would be a lot longer than it is now. There would still be a set of units where T[sup]2[/sup] = a[sup]3[/sup], and that set of units would be very natural to use in the modified solar system, but they would be different units than those used in our own Solar System.
Basically, you pick a planet (like, say, Earth, but it could be any of them). Define that planet’s average distance from the sun (whatever it is) as your length unit, and define the planet’s orbital period as your time unit. Then, in those units, you’ll have T[sup]2[/sup] = a[sup]3[/sup], for all other objects in that solar system.
Actuall the law that the square of the orbital period is proportional to the cube of the radius (for circular orbits) holds true. But the period isn’t always the same – the constant of proportionality depends upon the mass of the sun.
The reason this works is that the gravitational attraction of the sun depends on how far away you are from the sun. So every object as far away from the sun as the earth feels the same attraction to the sun. So for this object to stay in orbit around the sun it has to have a vector perpendicular to the sun, exactly equal to the vector towards the sun. This is what causes a circular orbit. The earth is constantly falling towards the sun at a constant rate, but since it has a sideways vector it is constantly missing.
If you decreased on increased the sideways vector you’d change the earth’s orbit from a near circular orbit to a more elliptical orbit. And if you did it enough, you change the orbit into a parabolic orbit (the earth would hit the sun) or a hyperbolic orbit (the earth would escape into interstellar space).
This Kepler’s Law Calculator gives the formula for the orbital period of a 2-body system as:
p^2 = a^3 / (M1 + M2)
Where:
P = orbital period in years
a = separation in Astronomical Units
M1 and M2 = masses of the 2 bodies in relation to the Sun’s mass
Actually, this is not quite complete. For example, assume that you have a circular orbit and you provide a boost to the planet’s orbital velocity - this would increase the semi-major axis of the orbit, and indeed make the orbital period longer. This is the basis for a variety of orbital transfer maneuvers, e.g. a Holman transfer “A spacecraft’s apoapsis altitude can be raised by increasing the spacecraft’s energy at periapsis.”
Just to make sure everyone caught that: If you speed up a planet, you will cause the period to increase, not decrease. In case anyone thought that Schuyler made a mistake.
Right, which means if you speed up the planet it will take longer to go around the sun. While if you slow it down it will go faster.
“I’m sorry, it seems we will arrive to early. We must increase speed.”
Actually Kepler’s Third Law is that the ratio of the orbital periods of two satellites of the same primary is the 3/2 power of their mean distances from the primary or t[sub]1[/sub]/t[sub]2[/sub] = (r[sub]1[/sub]/r[sub]2[/sub])[sup]3/2[/sup]. For any given primary the time for one orbit of a satellite is a constant times the 3/2 power of the distance to the primary. The constant depends upon the mass of the primary.