# Kepler law expression in other units [closed]

Consider the Kepler law : $$P^2=\frac{4\pi ^2 a^3}{G(m_1+m_2)}$$ I am asked to express this law in astronomical unit, solar mass and years. The solution of this problem states that in this case $$G=4\pi^2$$ and the Kepler law can be written as $$P^2=a^3$$ (in the solar system) and I really don't understand why.

As strange as $$G=4\pi^2$$ may seem, yes, it works if you use the right units for all variables. Let's have $$T$$ be in years; $$a$$ in AU (astronomical units); and $$m_1$$ and $$m_2$$ in solar masses ($$M_\odot$$).

In a metre-kilogram-second system of units, we use $$P^2=\dfrac{4\pi^2a^3}{G(m_1+m_2)}$$ with little trepidation.

Since between the two masses, of any Solar System planet (say $$m_1$$) and of the sun (say $$m_2$$), the sun's mass is much bigger, we may write $$P^2=\dfrac{4\pi^2a^3}{Gm_2}.\tag{1}$$ Note that $$m_2$$ is just 1 $$M_\odot$$.

Looking at the constants $$4\pi^2$$, $$G$$ and $$M_\odot$$, we can tell that, for any pair of bodies each orbiting the sun, $$\frac{P_1^2}{P_2^2}=\frac{a_1^3}{a_2^3}\tag{1}$$

Now, we don't want metres, kilograms and seconds; we want years (y), AU and $$M_\odot$$.

• But what's a year? It's the time taken for Earth to travel one round around the sun. That is, $$P_\text{Earth}=1$$ y.
• What's an astronomical unit? It's a currently fixed length1 meant to be roughly the distance between Earth and the sun. We may take $$a_\text{Earth}=1$$ AU.

Suppose we substitute our nice values into the denominators in (2): $$\frac{P_1^2}{1}=\frac{a_1^3}{1},$$ where subscript 1 denotes any body (including just Earth again) orbiting the sun.

Eh? Wait! We win! We have $$P^2=a^3$$ (again, supposing that the correct units are used).

But since (1) is true, $$\dfrac{4\pi^2}{G}=1$$ or $$G=4\pi^2$$, as you said the solution tells us. Technically, using dimensional analysis on (1), $$\text{y}^2=\frac{\text{AU}^3}{(\text{y}^{-2}\cdot\text{AU}^3\cdot M_\odot)\cdot M_\odot}$$

So, $$G=4\pi^2\text{ y}^{-2}\cdot\text{AU}^3\cdot M_\odot$$

Perhaps have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html (section 'The Law of Periods').

1. 1 AU = 149597870700 m

• Please note that policy on homework-type questions is to generally not provide a complete answer, but (where appropriate) to provide hints. – StephenG Apr 24 at 17:07