# How big to have objects orbit?

Maybe not this one: how big does an object (in space, presumably) have to be to have another object actually be so gravitically attracted to it to go into orbit, or fall (should I say “be attracted to”) the surface of the larger object?

Is it a size relationship? For example, say the big object is the size of those very large beach balls (4-5 feet in diameter) and the smaller object is a grain of sand. Would the grain of sand start to orbit the beach ball when it came close enough, or even fall to the surface of the beach ball?

Or is it a density relationship? The beach ball being made of lead, perhaps.

Just when does gravity start manifesting its power? Can an astronaut go into orbit (by himself, not in a ship) around a tiny asteroid, say, 1 mile in diameter?

These questions actually keep me up at night.

Orbiting can happen at any scale at all, as I understand it. For your lead beachball, it would be at a very small scale. Maybe on the order of grains of sand or dust motes. But it would happen, as the law of gravity doesn’t differentiate between the large and the small.

Orbiting is a special case of falling. If you fall straight down, you have no momentum to keep you from hitting the ground (or the beachball). If you fall with some horizontal momentum, like you’ve been launched from a cannon, you won’t hit the dirt quite so soon because you must first expend your momentum. If you fall with enough horizontal momentum (and enough is dependent on how strong gravity is, which is a function of how massive the two bodies are), you never hit ground. You keep going in an elliptical orbit, not having enough momentum to break free from gravity and go in a straight line off into space but having too much momentum to be drug down to the ground. You have entered freefall; that is, you are freely falling into the gravity well but not actually hitting anything. Freefall feels exactly like zero-gee, which is why people in freefall on, say, the Space Shuttle can float around and shoot Tang into little round globules to suck up.

So, yeah, I suppose molecules could orbit one another.

well, size and density. It’s based soley on Mass. the more mass an object has the more gravity it will exert. I give ya any calculations as i’ve never taken a physics class, but it takes quite a bit of mass to exert enough force to effect another object measurably. Your lead beachball most definitely would not effect a grain of sand. A hydrogen atom maybe? but i doubt even that. To put things in perspective earth iirc is many quintillion tons.

Ok, after reading derleth’s post I was hit with an epiphany of how stupid and uneducated in physics I am. ignore me!

In practice, you can only have a stable orbit when the gravitational pull between the two objects is strong compared to other forces acting on those bodies. The moon, for example, experiences more pull from the earth than from the sun, but if you move the moon farther away from the earth the gravitational pull becomes weaker and the orbit becomes less stable. Beyond some limit it won’t orbit around the earth, but instead it will be in an independent orbit around the sun. Similarly, if you were on the Space Shuttle and threw out an empty Coke can, the can will mostly see the gravitational pull of the earth, not the Shuttle. So the can will not orbit the Shuttle, it will orbit the earth independent of the Shuttle.

The force of gravity between two objects can be calculated from F = (G x m x m’) / r[sup]2[/sup]. G is the gravitational constant and is equal to 6.67 x 10[sup]-11[/sup] newton meters squared per kilogram squared. F will be in newtons if m and m’ are in kilograms and r is in meters.

Although gravity can happen at any scale, at some point, the electromagnetic force becomes more dominant. For example, although every atom in your body exerts a gravitational force on every other atom, you are being held together by the electromagnetic force.

Ever heard of space dust?

In classical quantum theory, electrons orbit protons because of the electromagnetic force that the protons exert on them. Due to the quantization of angular momentum, the tightest such an orbit can be is 5.3×10[sup]-11[/sup]m. This is called the Bohr radius. Now, the gravitational force, though there, is much weaker. If gravity were the force binding electrons to protons, the tightest orbit would have a size of 1.2×10[sup]29[/sup]m, or about 1000 times larger than the universe.

You gotta be careful talking about molecules, in which other, much stronger, forces than gravity come into play. Let’s assume there are no significant electrical (or other) forces. In that case, the only thing that matters is mass and distance to determine an orbit. We usually assume that the smaller body is orbitting the larger one (it becomes kind of moot when they are of similar size). With that convention, any object can orbit any other object. The smaller the orbitting object is, the faster it has to go to orbit (so it doesn’t crash into the larger object) or the farther out it has to be. I think it was Kepler who figured out that orbits were elliptical, even though he didn’t have Newton’s law of gravitation. Once you have the gravitational law, then it’s pretty easy to solve for orbits (which are elliptical). FYI, that law is F=GMm/r^2 (G=universal gravitational constant; M,m = the masses of the 2 bodies; r = the distance from the center of masses of the two bodies; F= force).

That can’t be right. Care to show your work, young man?

Just on a purely practical note, there are several known asteroids with “moons” (smaller asteroids orbiting bigger ones).

Scr said: “So the can will not orbit the Shuttle, it will orbit the earth independent of the Shuttle.” (Sorry, don’t know how to quote previous posts.)

But will it not orbit the Shuttle as they both orbit the Earth? And would not a grain of space dust orbit the can orbiting the Shuttle? (given sufficient distance, of course.)

I suppose what I meant was that both objects would presumably be outside an extra-gravitational force, but I see now that that is impossible, as even galaxies orbit around other galaxies.

In fact, I wonder if there is actually any point in the universe that is actually completely gravity-free.

If so, I would like to visit it.

[ul] NO[sub]there isn’t[/sub] [/ul]

Derleth described orbit pretty well: It consists of falling while moving sideways, so that you never actually hit.

Gravitation is an essential property of matter. So, all things exert some pull on all other things. It is certainly possible for a spacesuited astronaut to orbit a small asteroid. However, the smaller the objects involved, the thinner the tolerances. It would be very easy to push the astronaut just a little too hard, and he’d leave orbit. Asteroids have a very low escape velocity.

Also, while it’s conventional to treat the larger object as if it’s stationary, with the smaller object doing all the orbiting, it’s actually a mutual effect. The Earth is orbiting the Moon, the Sun is orbiting the Earth, etc.

Harmonix, it actually doesn’t take too much mass to exert a measurable force. It just requires a carefully designed apparatus to do the measurement. The first experiment to measure the universal gravitational constant used something very much like a lead beach ball. Here’s a link showing a more recent device, and you can see that the masses involved aren’t huge

Okay. Assume a circular orbit and let:

M[sub]p[/sub] = Proton mass
M[sub]e[/sub] = Electron mass
v = speed of electron in orbit
r = orbital radius (what we want to solve for)
F = gravitational force on electron from proton
L = angular momentum of electron

F = GM[sub]p[/sub]M[sub]e[/sub] / r[sup]2[/sup]
In a circular orbit: F = M[sub]e[/sub]v[sup]2[/sup] / r
This gives: GM[sub]p[/sub] = vr

L = M[sub]e[/sub]vr
Assume quantized angular momentum: L = hbar
This gives: M[sub]e[/sub]vr = hbar

Combine the two bolded equations and eliminate v to get: r = hbar[sup]2[/sup] / (GM[sub]p[/sub]M[sub]e[/sub][sup]2[/sup]) = 1.2 × 10[sup]29[/sup]m.

Truth to be told, though, I cheated. I didn’t derive this whole thing the first time around. I just replaced e[sup]2[/sup] / (4pi ep[sub]0[/sub]) with GM[sub]p[/sub]M[sub]e[/sub] in the expression for the Bohr radius.

Holy Christ, let alone the math, I’m just amazed you figured out the HTML to write those equations =+)

If you mean it feels like zero G to human perception when falling (or orbiting) Earth, then I concur.

If you mean that it feels like zero G in all ways exactly, then, no. Sensitive enough instruments can detect what humans can’t… the differences in tidal gravitational forces between the part of your body furthest away from the object you’re falling towards (or orbiting) from the gravitational force on the part of your body closest to the object you’re falling towards (or orbiting).

That’s why satellite “zero G” is often called microgravity, rather than “zero G.”

Of course, if you wish to orbit (or fall into) certain black holes where the gravitational tides will tear your body apart… then you just might conclude that freefall feels quite differently from zero G.

## Peace.

Ground control to Major Tom.

moriah, I used the word `feels’ to describe human perception, not what we can measure. As the word is usually used, in fact. Although microgravity is the more correct term, I think a human being would be very hard-pressed to tell the difference.

And I, too, concur that if you’re being disassembled by tidal forces, it will not feel like zero-gee.

And about molecules orbiting one another: Yes, I ignored electromagnetism. I assumed two electrically-neutral molecules at precisely the right distance and precisely the right velocity. Implausible? Yes, most likely. Impossible? No, not if Newton’s work means anything at all. (Of course, I had the presence of mind not to suppose two neutrons orbiting one another. The strong force is not to be ignored.)

Now I have a small question: Was it Newton who posed the famous cannon thought experiment that lead to the idea of orbiting being related to gravity? I can think of the image very well, but I can’t recall who drew it.