1. Nov 25, 2009

NCStarGazer

Looking at the gravitational equation

F=G*(M*m)/r^2

and centripetal force

F = ((V^2)/r)*m

If you set the two equal and solve for G you get:

G = ((V^2)*r)/M

Substituting (4*pi^2*r^2)/T^2 for V^2 you now have

G = (4*pi^2*r^3)/(M*T^2)

With solution for G, look at Kepler's law with Newton's update,

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

Substituting the G solved for into Kepler's equation and consider working with a perfect circle, r will equal a and T will equal P.

Now, once you substitute in the solved G, simplify...

You Get

(M+m)*P^2 = P^2*m

This is obviously not true!

Please let me know where the error is in my observation.

Thanks!

2. Nov 27, 2009

jmb

No! The law is $MP^2 = 4\pi^2a^3/G$ (see here). In which case there is no contradiction (hardly surprising given the above form of Kepler's Law is derived from Newton's law of gravity).

3. Nov 27, 2009

NCStarGazer

Well... I am confused, as the link you reference has 4π2a3 = P2G(M + m) under "Conversions for Unknowns", which is the same as what I referenced (M+m)*P^2 = (4*pi^2*a^3)/G.

4. Nov 27, 2009

D H

Staff Emeritus
A bit out of order. jmb's response first.

Sorry, jmb, but NCStarGazer has the right expression.

$$\tau^2 = \frac{4\pi^2}{G(M+m)} \,a^3$$

The source of NCStarGazer's error:
You are assuming that the r in Newton's law of gravity and the r in the centripetal force equation are one and the same. They aren't. Both objects are orbiting their center of mass. Given that the distance between a pair of objects of mass m and M is r, the distance from the object with mass m to the center of mass is

$$r_{cm} = \frac M{M+m}\,r$$

The centripetal force needed to sustain a circular orbit about the center of mass is thus

$$F=\omega^2 r_{cm} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{M}{M+m}\,r$$

Combining this with the gravitational force yields

$$\frac {GMm}{r^2} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{M}{M+m}\,r$$

Solving for the period,

$$\tau^2 = \frac{4\pi^2}{G(M+m)}\,r^3$$

5. Nov 27, 2009

NCStarGazer

D H

Thanks, I see how the center of mass can bring clarity. In working with your post I am still having trouble getting

$$\frac{GMm}{r^2}= \left(\frac{2\pi}{\tau}\right)^2\frac{M}{M+m}\,r$$

To yield your answer, I placed it in Maple and it always returns a version that has Gm(M+m); not G(M+m) in denominator ( I cannot find how to reduce the Gm to be only G), I did it by hand and I came up with the same. Did I miss something or was there a typo or such?

Appreciate the help!

6. Nov 27, 2009

D H

Staff Emeritus
A typo. ω2r has units of acceleration, not force. Restarting with the equation for the center of mass,

$$r_{cm} = \frac M{M+m}\,r$$

The centripetal force needed to sustain a circular orbit about the center of mass is thus

$$F=m\omega^2 r_{cm} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{Mm}{M+m}\,r$$

Combining this with the gravitational force yields

$$\frac {GMm}{r^2} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{Mm}{M+m}\,r$$

Solving for the period,

$$\tau^2 = \frac{4\pi^2}{G(M+m)}\,r^3$$

7. Nov 28, 2009

jmb

Ughh, my bad. I read this when I was very tired and tried to answer the question too quicky (something I really shouldn't do!).

I saw NCStarGar's use of $r$ in the centripetal and gravitational force equations and assumed he was making the assumption that $M>>m$ and ignoring centre of mass issues. Of course had I taken on the significance of what he/she meant by "Newton's update" I would have realised that wasn't the intent...

Sorry NCStarGazer and thanks to D H for spotting this. Very embarrassed now!