# Centre of Mass and tangential velocity star and planet

We have two objects - one massive star, and one planet which has a considerably smaller, but non-negligible mass, in a circular orbit about a common centre of mass. Using equation

$F$=$GMm$/$r^2$ where M is the mass of the star, m is the mass of the planet, and r is distance from the centre of mass of the two objects.

We can work out the velocity of the star about the centre of mass of the two objects by equating F to the centripetal force. This will give us

$v^2$=$GM$/$r$

This seems to indicate that the tangential velocity of the star is much greater than that of the planet, because as r gets smaller v becomes greater, and r is so much smaller than the star. What I understand is that the period of the two objects is equal, and so the star travels much slower to complete a orbit which has a far smaller distance travelled. This makes sense to me, however I want to know what's wrong with my maths, or perhaps my notation.

It seems that you are mixing up the distance between the planet and the star and the distance between the star and the centre of mass (both are called $r$ in your derivation).

Try to use $r = r_M + r_m$, where $r$ is the total distance between planet and star, $r_M$ is the distance between the star and the combined centre of mass,and $r_m$ is the distance between the planet and the combined centre of mass. You will need to use that the centrifugal force is given by $F = \frac{M v^2}{r_M}$.

Hope this helps!

• In the formula for newton's law of gravitation, is r not the distance from the star to the combined centre of mass? or must it be from the star to the planet? and why? Thanks! – inya Mar 27 '15 at 20:54
• got it, forgot that newton's law of gravitation meant that it was the separation between the two planets, not from planet to com – inya Mar 27 '15 at 21:10

First of all, it appears that you assume the centre of mass of the two-body system lies in the middle of the path between these bodies. This, however, isn't true. The centre of mass will lie closer to the star because that is where the majority of the mass is situated. If we define $r_M$ as the distance between the star and the c.o.m., and $r_m$ for the distance between the planet and the c.o.m., using the definition of c.o.m from this wikipedia article, http://en.wikipedia.org/wiki/Center_of_mass, it turns out that:

$$r_M = \frac{m}{M} r_m$$

i.e. the c.o.m. lies closer to the sun.

Now note that, while both bodies travel in circular orbits, they both have orbits of different radii: $r_M$ and $r_m$ for star and planet respectively. This means they have different centripetal force expressions. For this system to still work, however, we need the angular velocity, $\omega$ to be the same for both bodies: this way each body completes its orbit at the same time, which makes intuitive sense.

The gravitational force on both bodies is the same, by Newton's 3rd law. The centripetal force expression for the star is $F = Mv_M^2/r_M$ which can also be expressed as $F = M\omega ^2 r_M$. For the planet, it is $F = m\omega ^2 r_m$. If we can equate each centripetal force expression to the gravitation force, this means the centripetal force expressions themselves are equal.

Therefore:

$$M\omega ^2 r_M = m\omega ^2 r_m$$

However, we know from earlier, by definition of c.o.m. that:

$$r_M = \frac{m}{M} r_m$$

Therefore, it turns out that $\omega$ is constant all along. Happy days!

• It appears I wasn't clear in my original question - I knew the com was much closer to the star, and r from my two equations was actually the distance from the star to the combined centre of mass. I understand why w remains constant, and the speed can be derived from that - however, i can't derive the speed from equating centripetal force to gravity. Mv^2/r = G Mm/ r^2, where r is distance from star to com. M is mass of star and m is mass of planet. Where have I gone wrong? Thanks for your answer. – inya Mar 27 '15 at 20:53