I need to know how Kepler's third law formula is used...?

Question

I have the formula of a cubed/p squard=c I have the revolution of a planet that is 6 years. I don't want the answer to the question I just need to know how to GET the answer. I can't figure out how the math works? This is for an college astronomy class...

Answer 1

t^2 = k r^3

a useful value of k is 1 (why?). don't worry about the units.

what's the problem? this is high school arithmetic.

Answer 2

You haven't said so, but it sounds like you're being asked to find the size of the orbit of a planet whose period is 6 years.  Right?

You start with this:

A^3 / P² = c

Well, they don't tell you how much the constant "c" is; but no worries.  Since "c" is constant, it's the same "c" for EVERY planet, so plug in some known values for a known planet; namely the earth:

(A_earth)^3 / (P_earth)² = c

(1 au)^3 / (1 year)² = c

("1 au" stands for "1 astronomical unit", which is the radius of the earth's orbit).

That means the value of "c" must be (1 au)^3 / (1 year)²  (constant for all planets in the solar system).

Now apply it to your mystery planet:

(A_planet)^3 / (P_planet)² = c

But we said c = (1 au)^3 / (1 year)², so:

(A_planet)^3 / (P_planet)² = (1 au)^3 / (1 year)²

And the problem says that P_planet = 6 yrs, so:

(A_planet)^3 / (6 yrs)² = (1 au)^3 / (1 year)²

Solve for A_planet:

(A_planet)^3 = (6 yrs)² × (1 au)^3 / (1 year)²

= 36 × (1 au)^3

Take cube root of both sides:

A_planet = 1 au × cuberoot(36)

Answer 3

You just solve for whichever of the three variables you don't know.

P = 2 pi sqrt{ a^3 / (GM) }

a = { GM P^2 / (4 pi^2) }^(1/3)

M = 4 pi^2 a^3 / (G P^2)

First you convert the period from years into seconds, because this form of Kepler's third law is MKS (meter kilogram second units).

P = 6 years = 189345600 seconds

M = the sun's mass = 1.99E+30 kilograms

The product of the gravitational constant and the sun's mass is

GM = 1.32712440018E+20 m^3 sec^-2

So you solve for the semimajor axis, a, using the middle of the three formulas, above.

a = { [1.32712440018E+20 m^3 sec^-2] (189345600 sec)^2 / (4 pi^2) }^(1/3)

a = 4.939551E+11 meters

One AU = 1.49597870691E+11 meters

a = 3.301886 AU

There's a shortened version of Kepler's third law that works with units appropriate for sun-orbiting objects. The unit of mass is the sun's mass, and since the mass involved is always equal to one you don't need to put it in the shortened equation explicitly. The unit of time is years. The unit of distance is astronomical units.

P = a^(3/2)

a = P^(2/3)

a = 6^(2/3) = 3.301927 AU

Well, it's close anyway.