What we have:
\[ P \: \alpha \: M\cdot a \] where P is period, M is mass, and a is the separation of a two-body gravitational orbit.
What we know:
Because gravity is involved, we know G will be used, which is \( G = 6.67 x 10^{-12}\frac{cm^3}{gs^2}\).
We also know the units:
Period: T (time), M: M (mass), a: L (distance) and finally G: \(\frac{L^3}{MT^2}\) (a combination of the three).
So we can set up a dimensional analysis problem with only the exponents to work out:
\[ P^w=M^x\cdot a^y\cdot G^z \rightarrow T^w=M^x\cdot L^y\cdot (\frac{L^3}{MT^2})^z\]
For the units of time (T) to balance out on both sides of the equation, w must equal 2 and z must equal -1. With this inversion, for the units for distance to work out (L), y must equal 3. And finally, for mass to cancel (M), x must equal -1.
In conclusion, w=2, x=-1, y=3, and z=-1, which when put back into the first equation gives:
\[ P^2=M^{-1}\cdot a^3\cdot G^{-1} \]
which cleans up to:
\[ P^2=\frac{a^3}{M\cdot G} \]
All done!
P.S. Because I do not live on a desert island, I looked up the formula for Kepler's law:
\[ P^2=\frac{4\pi^2 a^3}{MG} \]
and I was only off by a constant!
Many thanks to Sean and Eden for working on this worksheet with me.
This is nice walk-through of the problem. And you missed a \(\pi\). They are sneaky. They are fun because radians are technically dimensionless (you can take the \(\sin{}\) of unit), so that are hard to find with dimensional analysis. Can you think of why it's \(4\pi^2\) or (hint, hint) \( (2\pi)^2) \)?
ReplyDeleteI fixed the pi. This makes sense because we are finding period over an entire circle, so the equation is really \[ \left( \frac{P}{2 \pi} \right)^2=\frac{a^3}{M\cdot G} \] which simplifies to the final equation.
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