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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.

Qmechanic
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Dicordi
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  • Sorry, that should be $GM=4\pi^2$, where $M$ is the solar mass. So you're being told to use solar masses as the mass unit, i.e., the mass of the Sun = 1. FWIW, $Gm=\mu$ is known as the standard gravitational parameter. Also see https://physics.stackexchange.com/q/546062/ – PM 2Ring Apr 24 '20 at 12:01

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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

subdermatoglyphic
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