Orbital period?

Question

Is it possible to find the orbital period of a planet just by knowing its semimajor axis?

What.. By using

period = (semimajor axis)^3/2??

so if I measure the semimajor axis in units Km, what units will the period be given in?

Answer 1

Yes. Use Keplers' Third Law of Planetary Motion which states the squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axes (the "half-length" of the ellipse) of their orbits.

IF
P = orbital period of planet
a = semi-major axis of orbit

The expression P^2 : a^–3 has the same value for all planets in the Solar System as it has for Earth. When certain units are chosen, namely P is measured in sidereal years and a in astronomical units, P^2 : a^–3 has the value 1 for all planets in the Solar System.

This is also known as the Harmonic Law.

You can cube the figures in the left-hand column and square the figures in the right-hand column, to satisfy yourself that it works.

Semi-Major Axis (AUs) Orbital period (years)

Mercury 0.387 0.241
Venus 0.72 0.615
Earth 1.00 1.00
Mars 1.52 1.88
Jupiter 5.20 11.86
Saturn 9.54 29.46
Uranus 19.22 84.01
Neptune 30.06 164.8

HERE IS A LITTLE TEST FOR YOU TO SEE IF YOU UNDERSTAND IT

Here are five Trans Neptunian Objects with the length of their semi-major axes in AUs:

(55636) 2002 TX300 43.102 AU
20000 Varuna 43.129 AU
(136108) 2003 EL61 "Santa" 43.339 AU
50000 Quaoar 43.376 AU
(136472) 2005 FY9 "Easterbunny" 45.64 AU

Here (jumbled up into a different order) are the Oribtal periods of these 5 objects:

285.4 years
282.84 years
309.87 years
285.97 years
283.2 years

Which Orbital period belongs to which object?

EDIT

To the questioner: by all means get the semi-major axis in kilometres or miles as that is how it is often quoted, But first convert it to Astronomical Units (AUs), then the sums are easy and the orbital period in years just tumbles into your lap.

To the fourth contributor:

If you had read our pieces more carefully, you would have noticed that only one of us used an equals sign, and two of us said the ratio was proportional. So why are you misrepresenting what two of us say and tarring us all with the same brush?

I am well aware that in a different stellar system from our solar system, different numbers would apply, as the other star would not have the gravity or mass of our Sun and a different constant would be needed. But as there are, I think, only 6 stars with 3 or more known exoplanets at present, it seemed a little academic to explore that topic very far. Anyone wanting to calculate orbital periods is going to want to do it in the Solar System!

Answer 2

Yes. The squares of the orbital period is directly proportional to the cubes of the semimajor axis.

Kepler's Third Law of Planetary Motion, which correlates a planet's orbital period with a planet's distance from the Sun, can be summed up by the simple equation: T2 = D3. In this equation, T = the time of the planet's orbital period in Earth years and D = the planet's mean distance from the Sun in astronomical units (AU). One AU = the mean distance between the Earth and the Sun.


Kepler's Third Law governs all bodies orbiting the Sun, including asteroids and comets. That's because the Sun claims more than 99.8% of the total mass of the Solar System. If Solar System bodies were substantially more massive, Kepler's Third Law -- as stated -- would not be sufficient.

Let's presume that this body happens to have significant mass -- 1/10th the Sun's. Then the equation must be modified M = the mass of the Sun and m = the mass of the secondary body:

(M+m) x T2 = D3


When you use this equation, remember to have the orbital period in Earth years, the distance in astronomical units and the mass in solar masses.

Answer 3

Yes. Kepler's 3rd law. Period (in years) squared equals semimajor axis (in AU) cubed, or P^2=A^3. One AU is the semimajor axis of the Earth's orbit.

Answer 4

NO! The above posters are incorrect....

P^2 is only inversely PROPORTIONAL to A^3.....NOT equal to it.

Therefore more info is needed......namely the value of G (gravitational constant) and the mass of the sun, M.

The complete Kepler equation is ......A^3/P^2 = GM/(2pi)^2

In general you need G and M, along with A in order to predict a particular planet's orbital period; (however, it is usually measured)....Note, the A^3/P^2 ratio for ANY planet is a constant.

Thus, IF you KNOW ANOTHER planet's A and period, then you can compute any planet's period (knowing only its semi-major axis) since the ratio of each of them is equal to the same constant, namely....GM/(2pi)^2 ...(BUT YOU MUST HAVE THE VALUE OF THAT CONSTANT, either from knowing the A/P ratio of another planet OR by knowing the value of G and M independently).

Also, IF you can estimate the distance (semi-major axis) and can measure the period, it is possible to determine the mass of the sun...which (BTW) is exactly what Kepler did!

Answer 5

ok @ first i pronounced 24 hrs yet i assume its 23 hrs fifty six minures - the comparable era using fact the earth that way a Geo sych satellite tv for pc continually keeps to be over your head and sends you your cellular telephone sign! keplers 2d regulation is used to discover the radius (or era) of orbit