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I've been trying to google the answer to this question, but have had no luck, so I'm asking this question here. Let's say the origin is at (0, 0), body 1 with mass m1 is at (-r1, 0), and body 2 with mass m2 is at (r2, 0). How can I calculate the initial velocities v1 and v2 for the bodies so that they trace circular orbits around each other?

The reason I'm asking is because I've only been able to get the answer for the specific problem where m1 = m2, but I can't explain it very well, and because I can't explain it very well I can't generalize it to bodies of different masses.

Thanks for any assistance!

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up vote 2 down vote accepted

I will assume the phrase "circular orbits around each other" means that you are looking for a solution where the two bodies remain constant distance apart.

The easiest thing to do is to work in a coordinate system adapted to the centre of mass. The total separation between the two masses is $R = r_1 + r_2$. The centre of mass sits at $(R_0,0)$ where $R_0 = (-m_1 r_1 + m2 r_2)/(m_1 + m_2)$. Without loss of generality, we'll assume that the centre of mass remains immobile (else we can subtract the velocity of the centre of mass from both bodies and use the fact that Galilean motions leave the mechanics invariant).

So the distance from $m_1$ to the centre of mass is $|-r_1 - R_0| = R m_2 / (m_1 + m_2)$, similarly the distance from $m_2$ to the centre of mass will be $R m_1/(m_1 + m_2)$.

Now, let us write down the equations of motion.

  • The assumption that the centre of mass remains immobile implies that the total momentum is 0. This means that $m_1 v_1 + m_2 v_2 = 0$. Furthermore, that $R$ is constant requires that the initial velocities be such that the two masses start out moving perpendicular to the $x$ axis. So hereon we can assume that $v_1, v_2$ are scalars, which represent the initial speed in the $y$ direction for the masses respectively.

  • Relative to the (immobile) centre of mass, the first mass is travelling in a circular orbit of radius $R m_2 / (m_1 + m_2)$. This requires force $ m_1 v_1^2 / (R m_2 / (m_1 + m_2) ) = v_1^2 \cdot \frac{m_1 (m_1 + m_2)}{R m_2}$. We have also a similar expression for the force acting on $m_2$.

  • The gravitational attraction between the two masses is $G m_1 m_2 / R^2$.

So now we solve the system

$$ m_1 v_1 + m_2 v_2 = 0 $$ $$ \frac{G m_1 m_2}{R^2} = v_1^2 \frac{m_1 (m_1 + m_2)}{m_2 R} = v_2^2 \frac{m_2(m_1 + m_2)}{m_1R} $$

Which gives

$$ v_1 = \sqrt{ \frac{G m_2^2}{R(m_1 + m_2)} }, v_2 = \sqrt{ \frac{G m_1^2}{R(m_1 + m_2)} }$$

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This makes sense. It's exactly what I wanted. Thanks! – bob Nov 7 '10 at 22:04

Silly question, but have you read the Wikipedia page on the two-body problem ( If I understand your problem correctly, then it seems that it is fairly well treated on that page.

You say you want to determine the set of initial vectors $v_1$ and $v_2$ such that the two bodies trace circular orbits around each other. If $x_1(t)$ and $x_2(t)$ denote the positions of the bodies at time $t$, then do you mean that you want the distance $|x_1(t) - x_2(t)|$ between the two to be equal to some fixed radius $r$? If so, then this seems to be treated in the "Displacement vector motion" section of the Wikipedia page (if the radius $r$ is constant, then its derivative w/r/t time is ...?).

The WP page seems to solve two one-body problems to arrive at a complete description of the system of solutions. In contrast, you only ask for a partial description of the system (i.e. what inital vectors give circular orbits). An alternative solution in your case could be to redefine the origin of your coordinate system to be given by $x_1(t)$ at time $t$; then your job is just to determine which initial vectors give circular orbits of one mass in orbit around another fixed one.

The second method actually sounds like it's more instructive and fun to do than working through the WP page. I'd think of Isaac Newton and try the second method to start with, then turn to the WP page (or here) if I got stuck. Good luck.

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I saw the WP page and it wasn't explained clearly enough for me to understand it. – bob Nov 7 '10 at 22:03

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