# How do you travel in a circular orbit around a massive body?

I am trying to figure out how an object could achieve a perfectly circular orbit. Given a mass for the planet or other body the object is orbiting and a distance from the center of mass, how fast would the object have to be moving perpendicular to the center of mass?

My initial assumption was it would have to be travelling horizontally (relatively speaking) as fast as it was falling towards to center of mass. Is this a correct assumption? I guess I am confused because it is easy to calculate the acceleration towards the center of mass given the distance and mass of the body, but I have no idea how you would go about calculating the velocity necessary.

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Think about centripetal force: en.wikipedia.org/wiki/Centripetal_force –  Will Jul 24 '13 at 22:30
Are you considering a system in which the bodies are isolated from all other matter? If the answer to that is yes, then the frame in which the center of mass is at rest is inertial. Are you asking about observing the orbit in this frame? –  joshphysics Jul 24 '13 at 22:31
Yes, the system is completely isolated from all other matter. –  CoderTheTyler Jul 24 '13 at 22:32

The condition for staying in a circular orbit is the requirement for the centripetal force to be equal in magnitude to the gravitational pull. To be precise:

$$F_g=F_c,$$

$$mg=\frac{mv^2}{r},$$

where $F_g$ is the absolute value of gravitational force, $F_c$ the absolute value of centripetal force, $g$ the gravitational acceleration, $m$ the mass of the moving object, $v$ its tangential velocity and $r$ the distance from the center of the orbit. You can express the required velcocity from that equation, which yields:

$$v=\sqrt{rg},$$

which is independent of the mass of the object. Note that $g$ is not the gravitational acceleration near the earth's surface, but the acceleration the object experience due to the gravitational field at its current location. It is given by

$$g=G\frac{M}{r^2},$$

where $M$ is the mass of the body around which the moving object orbits and $G$ is Newton's gravitational constant.

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That seems to work very nicely! I ran a little simulation with it and the orbiting body stays practically the same distance away from the center of gravity. It also makes perfect sense now that I think about it. Thanks for the help! –  CoderTheTyler Jul 24 '13 at 22:46
I feel that this answer is misleading because it doesn't make clear that the gravitational acceleration depends on the distance between the two bodies and does not necessarily equal the gravitational acceleration close to the surface of the earth which is what the symbol $g$ commonly refers to. –  joshphysics Jul 24 '13 at 22:53
I will add a few lines to make it clearer. –  Frederic Brünner Jul 24 '13 at 22:56

An alternative approach: the orbit has two constants of motion: conservation of energy and conservation of angular momentum. The energy is the sum of the potential and kinetic energy \begin{align} E &= -\frac{GMm}{r} + \frac{1}{2}mv^2\\ &= -\frac{GMm}{r} + \frac{1}{2}mv_r^2 + \frac{1}{2}mv_T^2,\\ \end{align} where $v_r$ is the radial velocity and $v_T$ the tangential velocity component. The magnitude of the angular momentum is $$L = mrv_T.$$ Substituting $L$ into the equation for $E$, we get $$E = -\frac{GMm}{r} + \frac{1}{2}mv_r^2 + \frac{L^2}{2mr^2},$$ so that we can write $v_r$ as a function of $r$: $$v_r^2 = \frac{2E}{m} + \frac{2GM}{r} - \frac{L^2}{m^2r^2}.$$ For a circular orbit, $v_r$ is identically zero, which means that its derivative is also zero: $$\frac{\text{d}v_r^2}{\text{d}r} = -\frac{2GM}{r^2} + \frac{2L^2}{m^2r^3} =0,$$ so that $$r = \frac{L^2}{GMm^2}.$$ Therefore $$L^2 = GMm^2r = m^2r^2v_T^2,$$ in other words $$v_T = \sqrt{\frac{GM}{r}}.$$

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