# Homework Help: Formula relationship

1. Mar 13, 2005

### tony873004

Show that
$$4\pi^{2}/G = 1yr^{2} M_{Sun}/AU^{3}$$
and that therefore Kepler's 3rd Law reduces to
$$P^{2}=a^{3}/(m_{1}+m_{2})$$
if the period P is expressed in years, the semimajor axis a is expressed in AU, and the sum of the masses is expressed in solar masses.

$$G = 6.672*10^{-11}Nm^{2}/kg^{2}$$
$$N = kg m/s^{2}$$
$$1 year = 3.15581*10^{7}seconds$$
$$1 AU = 1.49598*10^{11}meters$$
$$1 Solar mass = 1.9891*10^{30}kg$$

$$\frac{4\pi^{2}}{6.672*10^{-11} \frac{N m^{2}}{kg^{2}}}=\frac {3.15581*10^{7}s^{2}*1.9891*10^{30}kg}{1.49598*10^{11}m^{3}}$$

Substitute $$kg m/s^{2} for N$$

$$\frac{4\pi^{2}}{6.672*10^{-11} \frac{kg\frac{m}{s^{2}} m^{2}}{kg^{2}}}=\frac {3.15581*10^{7}s^{2}*1.9891*10^{30}kg}{1.49598*10^{11}m^{3}}$$

in the left side of the equation, cancel kg's and combine m's

$$\frac{4\pi^{2}}{6.672*10^{-11} \frac{\frac{m^{3}}{s^{2}}}{kg}}=\frac {3.15581*10^{7}s^{2}*1.9891*10^{30}kg}{1.49598*10^{11}m^{3}}$$

re-write the equation

$${4\pi^{2} * 1.49598*10^{11}m^{3} = 6.672*10^{-11} \frac{\frac{m^{3}}{s^{2}}}{kg} * 3.15581*10^{7}s^{2}*1.9891*10^{30}kg$$

The m^3 on the right cancels with the m^3 on the left
the s^2's in the left side of the equation cancel with each other since one is in the denomanator of a fraction, and the other one isn't.
The kg's in the left side of the equation cancel with each other since one is in the denomanator of a fraction, and the other one isn't.

$${4\pi^{2} * 1.49598*10^{11} = 6.672*10^{-11} * 3.15581*10^{7}*1.9891*10^{30}$$

and when I compute this I get:

$$591702901743.965 = 4.1881622988912*10^{27}$$
and that's just not right!! Where did I go wrong?

Last edited: Mar 13, 2005
2. Mar 13, 2005

### Data

$$1 \mbox{year} = 3.15581 \cdot 10^7 \mbox{s} \Longrightarrow (1 \mbox{year})^2 = (3.15581 \cdot 10^7 \mbox{s})^2 \neq 3.15581 \cdot 10^7 \mbox{s}^2$$

similarly for $$\mbox{AU}^3$$.

3. Mar 13, 2005

### tony873004

Thanks, Data. That's the 2nd time you've bailed me out in the last 2 days.

I also made a small error in G. Should be $$6.673*10^{-11}$$, not $$6.672*10^{-11}$$

So this should change my final line before computing the numbers to:

$${4\pi^{2} * 1.49598*10^{33} = 6.673*10^{-11} * 3.15581*10^{14}*1.9891*10^{30}$$

which gives me
$$5.90589231677665*10^{34} = 4.1887900210583*10^{34}$$

They're not too different. At least they're in the same magnitude instead of being 16 magnitudes apart. I'm wondering if this is as good as it gets or if I made another error somewhere?

Any idea how to do the 2nd part of this question, how does this cause Kepler's 3rd law to reduce to $$P^{2}=a^{3}/(m_{1}+m_{2})$$ ?

Last edited: Mar 13, 2005
4. Mar 13, 2005

### Data

Almost. You forgot to square the 3.15581, and cube the 1.49598!

The second question is a really ambiguous one. In order to get it to that form, you must assume that both sides of the equation are unitless (otherwise, you keep the fraction $$\frac{\mbox{yr}^2}{\mbox{AU}^3}$$ on the right side of the equality to balance units). This in turn means you just assume the result for P is in years, etc., which is silly. Anyways, Kepler's Third Law is

$$P^2 = \frac{a^3}{m_1+m_2}\cdot \frac{4\pi^2}{G}$$

Just substitute in the equality that you proved in the first part, and see how far you can get~

5. Mar 13, 2005

### tony873004

Thanks, Data. I squared and cubed those numbers, and now the answers are equal. I got the second part too. I had to make a new value for G, which equaled 4pi^2, they cancelled out and I was left the equation I was asked to justify.

6. Mar 14, 2005

### Data

You set $$G = 4 \pi^2$$? With what justification?

7. Mar 14, 2005

### tony873004

$$G=6.673*10^{-11}N m^{2}/kg^{2}$$ , therefore
$$G=6.673*10^{-11}kg (m/s^{2}) m^{2}/kg^{2}$$

substitute in the conversions from mks to AU, Solar Masses, Years

0.00000000006673 * ((1 / 1.9891E+30) * (1 / 149598000000#) / (1 / 31558100) ^ 2#) * (1 / 149598000000# ^ 2 / (1 / 1.9891E+30) ^ 2)

and magically the whole thing equals exactly 4pi^2
Sorry, I didn't feel like TEXing it.

By converting all the units in G to AU, Years, and Solar Masses. Now G cancels out with 4pi^2
$$P^2 = \frac{a^3}{m_1+m_2}\cdot \frac{4\pi^2}{G}$$

reduces to

$$P^{2}=a^{3}/(m_{1}+m_{2})$$

Which is what the question asked me to do.

8. Mar 14, 2005

### Data

Ok, that's fine. Make sure you leave in the unit conversion term, though (ie. the magnitude equals the magnitude of $$4\pi^2$$, but there are still units hanging around that $$4\pi^2$$ does not carry). You really can't simplify it to the form in your question, as there's always a unitary conversion factor lying around. But whatever, I guess they want you to write it that way :)

9. Mar 14, 2005

### tony873004

Yeah, those units bugged me too. So I just dropped them all together 2 steps before my final answer. When the teacher talked about this one in class, everyone had a blank stare. But she hinted that this was the way to do it. At least I'll get effort points!!

Thanks for all your help, Data!

10. Mar 14, 2005

### tony873004

I asked in class today. The units do carry down to the 1.0 computed with 4pi^2/G. Then when you replace the 4pi^2/G in the Kepler's law with 1.0&units, the units cancel with the units in the rest of Kepler's law so that P^2 comes out in years.

11. Mar 14, 2005

### Data

I agree, but then the $$P, \ a, \ m_1, \ m_2$$ in your new equation aren't the same ones in the original equation (the new ones are unitless). Essentially the point is that $$\mbox{time} \neq \frac{\mbox{distance}}{\mbox{mass}}$$.