# Orbital speed for a circular orbit?

I wanted to look up the formula for orbital speed for a circular orbit on Wikipedia, and I found 2 formulas:

All bounded orbits where the gravity of a central body dominates are elliptical in nature. A special case of this is the circular orbit, which is an ellipse of zero eccentricity. The formula for the velocity of a body in a circular orbit (orbital speed) at distance r from the centre of gravity of mass M is $v = \sqrt{\dfrac{GM}{r}}$.

I found this weird, because this leaves out the mass of the body orbiting $M$. I would think that this would have some effect on the orbital speed. I figured that perhaps the radius of the orbit indicates the mass of the orbiting body through some physical relation, but I wasn't sure so I continued looked up stuff on Wikipedia and I found:

The relative velocity is constant: $v = \sqrt{\dfrac{G(M+m)}{r}}$.

This was the equation I intially expected to see, but now I'm confused because these are 2 formulas for the same situation, right? Or does the word 'relative' indicate a difference?

• Sep 25, 2013 at 23:47

I think these are two separate questions that should be approached separately.

1) "Why isn't $m$ in in the first equation?" The mass of a body does change the force acting on it. But the mass of a body also changes its acceleration. If you increase the mass of an object it feels a larger force, but it's also harder to move. The equation for gravitational force is $F = \displaystyle GmM/{r^2}$, while the equation for acceleration is $a = F/m$. Glue to two together and you get $a = GM/r^2$.

2) "Why is $m$ in the second equation?" Think about the moon and the earth. The earth is pulling on the moon, but the moon is also pulling on the earth! The two bodies actually orbit around their common center of mass. This is important for the relative velocity: we need to add how fast the earth is orbiting to how fast the moon is orbiting. That's why you have the $m$ term.

• So what is the difference between them? It seems as if you refute part 1 in part 2. In which situations do you apply eq. 1 and in which situations do you apply eq. 2? Sep 25, 2013 at 23:04
• @user30117, The orbit speed of the moon around the center of mass is part 1. If you were on the earth looking at the moon, you'd measure its speed as in part 2 since the earth is also moving and you have to take that into account. Sep 25, 2013 at 23:07
• Equation 1 is for the orbit speed around the centre of mass, equation 2 is for the orbit speed around the actual earth? That sounds logical, thanks. Sep 25, 2013 at 23:18
• I'd add that $v=\sqrt{GM/r}$ is a good approximation when $M>>m$, and is commonly used in rough calculations, even if it is strictly wrong for the situation. Mar 4, 2014 at 21:40

The relative velocity in a circular orbit is indeed: $v_\text{rel} = \sqrt{ \dfrac{G(m_1+m_2) }{ r_\text{rel}} }$

The relative velocity is the sum of the barycentric velocity of each body (the velocity of each body with respect to the inertial center of mass): $$v_\text{rel} = v_1 + v_2$$

$$v_1 = \frac{m_2}{m_1 + m_2} v_\text{rel}, \quad \quad v_2 = \frac{m_1}{m_1 + m_2} v_\text{rel}$$

Similarly, $r_\text{rel}$ is the relative separation of the two bodies.

$$r_\text{rel} = r_1 + r_2$$

Where $r_1$ and $r_2$ are the distances of each body from their mutual center of mass

$$r_1 = \frac{m_2}{m_1 + m_2} r_\text{rel}, \quad\quad r_2 = \frac{m_1}{m_1 + m_2} r_\text{rel}$$

The relative velocity is also given by: $$v_\text{rel} = \sqrt{\frac{G m_1}{r_2}} = \sqrt{\frac{G m_2}{r_1}}$$

The barycentric velocity of body 2 is then: $$v_2 = m_1 \sqrt{ \frac{G}{(m_1+m_2) \, r_\text{rel}} } = \frac{m_1}{m_1+m_2} \sqrt{ \frac{G m_1}{r_2} }$$

This is the actual velocity of body 2 in the inertial frame of the barycenter.

The first equation is a very close approximation since m (satellite's mass) << M (Earth's mass) so m can be ignored. The second equation is the mathematically correct one.

It depends on whether you're discussing artificial (or otherwise small) satellites vs. things like moons.

In the case of artificial satellites, M is so much larger than m that, for all practical purposes, M + m equals M, which makes the two equations equivalent and the "+ m" completely irrelevant.

For example, when considering the Earth and the International Space Station, the former is on the order of 10^19 times larger than the latter. That is, M + m = M out to around 19 decimal digits. Considering that your calculations will almost certainly not have 20 significant figures, you can consider M + m to be equal to M with no loss of precision at all.

The situation changes when you start discussing much larger things like moons. In the case of Earth's moon, for example, it's a little over 1% of the mass of Earth, so M + m is not equal to M for anything over 2 significant figures. Most useful calculations will need more than 2 significant figures.