My basic understanding of what achieves orbit is: The gravitational pull of the larger object (Earth) + the natural force of the mass object moving forward (The Moon) = orbit.

With this being true, how come there are different levels of orbit, like the Low Earth Orbit which I keep reading about? Doesn't the mass of the satellite dictate the height/location of the orbit?


2 Answers 2


Doesn't the mass of the satellite dictate the height/location of the orbit?

No, at least not to first order. There is some effect in the case of objects whose mass is close to that of the object it is orbiting, such as the Moon orbiting the Earth. I'll deal with that later. I'll focus at first on objects that are many orders of magnitude smaller than the object they are orbiting. For example, large as it is, the mass of the International Space Station is less than 10-19 Earth masses. The mass of the Space Station plays no role in determining it's orbit. The key to determining the orbit of some satellite about a primary (e.g., the Space Station about the Earth, or the Earth about the Sun) is the satellite's velocity with respect to the primary.

There is no need to invoke fictitious forces such as centrifugal force to explain why things orbit. The first thing to do to understand orbits is to look to Newton's first law of motion. A satellite's orbit is a curved path (i.e., not a straight line) about the primary. Since the path is curved rather than a straight line, there must necessarily be some net force acting on the satellite to make it follow that path. That force is of course gravitation. No other forces are needed.

Imagine an airless, non-rotating planet with a cannon that shoots cannonballs horizontally mounted atop the planet's tallest mountain. (This is Newton's cannon.) If you use but a small amount of gunpowder to fire a cannonball, the cannonball will appear to follow a parabolic arc, just as you learned as a kid. If you use more gunpowder you'll find that the path is no longer parabolic. It is instead a segment of an ellipse. If you use even more gunpowder, the ellipse will no longer intersect the planet's surface. The cannonball is in orbit. If you use just the right amount of gunpowder, the cannonball will orbit at a constant height above the planet. This is a circular orbit. (But note that a circle is just a special case of an ellipse.) With even more gunpowder, the orbit is once again elliptical, but now the top of the mountain is the closest point on the cannonball's trajectory about the planet. Finally, if you use more gunpowder yet, the cannonball will have enough energy to escape the gravitational clutches of the planet. In this case, the path is a parabola or hyperbola rather than an ellipse.

Suppose you can raise the cannon hundreds or even thousands of kilometers above the top of the mountain. As you do so, you'll find that the amount of gunpowder needed to have a cannonball orbiting circularly about the planet decreases. The circular orbit velocity is solely a function of the planet's mass $M$ and the radial distance $r$ from the center of the planet, $v_\text{circ} = \sqrt{\frac{GM} r}$, where $G$ is Newton's universal gravitational constant. You'll also find that the escape velocity similarly decreases with radial distance, $v_\text{escape} = \sqrt{2 \frac{GM} r}$.

Finally, what about the Moon? Unlike the Space Station, the Moon's mass compared to that of the Earth is not negligible. It is more than 1/100 of the Earth's mass. (To be more precise, it's about 0.0123 Earth masses, a conveniently easy number to remember.) The above expression for the circular orbital velocity is not quite correct. In this case, it's better to use $v_\text{circ} = \sqrt{\frac{G(m_1+m_2)} r}$, where $m_1$ and $m_2$ are the masses of the primary and the satellite.


Orbits are possible due to centrifugal force. The faster something is going (in angular velocity, which you can think of as "the speed at which it goes around something" or "the times it goes around a thing in a minute/day/year/whatever"), the stronger the centrifugal force and the closer to earth you can orbit.

Mass doesn't matter, or rather it cancels itself out. There's a mass term in the formula for centrifugal force: F = m * [d^2 r]/[d t^2] which means Force = mass times the angular velocity (the change in direction around something divided by time, just like normal velocity is the change of position divided by time), squared (meaning multiplied by itself). Angular velocity is often written as a lower-case Greek letter omega: ω. Thus, we can re-write the formula for centrifugal force as F = m * ω * ω. There's also a mass term (actually, two of them, one for each object, but for this purpose we treat the earth as fixed and are only concerned about the satellite's mass) in the formula for gravity: F = G * m1 * m2 / r^2 which means that the Force of gravity is equal to the Gravitational constant (you need to know this to determine the actual force as a number, but for now, just ignore it) times the mass of the first object (earth) times the mass of the second object (the satellite), all divided by the radius (the distance between the objects' centers) squared (multiplied by itself).

In order to orbit at a specific height (commonly called an "altitude"), your outward (centrifugal) force must equal your inward (gravitational) force. Thus, to orbit at height r (from the center of the earth) you solve this formula: m * ω^2 = G * m1 * m2 / r^2. Since the mass of the satellite is on both sides of the equation, you can simply divide both sides of the equation by that mass - it doesn't matter what it is! - and you have a simpler equation that gives the same result: ω^2 = G * m1 / r^2. If you plug in all the numbers into that, it will tell you how fast anything needs to be going around the earth in order to orbit the Earth at a given height.

Since gravity falls off rapidly as you go further out - something twice as far as you from the center of the earth experiences only one fourth as much gravity - stuff that is relatively close to earth (things in low earth orbit, such as the International Space Station) have to go around the earth really fast; the ISS (and everything else at the same height) does it in about 90 minutes. That's only a few hundred miles (or a few hundred kilometers) above the earth's surface. Many times that far away - about 26.2 thousand miles, or 46.2 thousand kilometers - stuff orbits in exactly one day, which means they always pass over the same spot on the equator every day. If they orbit over the equator directly, they don't appear (from those of us on earth) to move at all. This is useful for communication and TV satellites; you can point a fixed satellite dish at them. Much further out still, at 385 thousand kilometers, you have the orbit of the moon, which takes about 27.3 days (just shy of four weeks). These are just three examples of different orbital heights around earth.

You can demonstrate this easily yourself if you have a yo-yo or similar (really, any string with a weight on the end). Start with a long section of string, at least two feet long (60 cm) and spin it around your hand just fast enough to keep the string tight and have the weight make a full circle. Now shorten the string a bunch, to maybe eight inches (20 cm) and again, spin it fast enough to make a full circle. You should notice that, with the shorter string, you have to spin the object a lot faster to keep the string tight, but you can keep it tight at the different distances. The tension in the string stands in for gravity, here; if the string is loose, gravity would overcome your centrifugal force and crash your satellite. If you spin (orbit) so fast that the string is under too much tension and breaks, that's like going so fast you leave earth's orbit (though unlike a string, gravity never "breaks"; it's more like it would just stretch out, letting the satellite fly away and exerting a weaker and weaker pull to try and bring it back).

Hope that helps!

  • $\begingroup$ Thanks for the very comprehensive answer! Strangely if anything I would have thought the inverse was true about the centrifugal force, in that the faster you go, you are less subject to the pull of Earth's gravity. (Isn't this why pedalling on a bike keeps you upright, but being stationary becomes much more difficult to stay upright?) $\endgroup$
    – Ricky
    Commented Dec 12, 2015 at 0:15
  • 3
    $\begingroup$ There is NO reason to invoke centrifugal force to explain how things orbit. Downvoted. $\endgroup$ Commented Dec 12, 2015 at 1:36
  • $\begingroup$ @DavidHammen: I'm sorry, but I disagree. Centrifugal force is what makes things orbit (as opposed to crashing into the gravity well). You may as well say "There is NO reason to invoke buoyancy to explain how things float." In a stable orbit, gravity provides the centripetal force that balances the object's centrifugal force. Centrifugal force is easy to comprehend (and even observe) without much physics knowledge, at least for circular orbits, so I thought it would aid understanding. Is this an "it's just inertia, not a real force" complaint? Should I have invoked relativity? $\endgroup$
    – CBHacking
    Commented Dec 12, 2015 at 7:06
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    $\begingroup$ @Ricky: Speed doesn't actually matter for gravity, just the objects' masses and the distance between them (OK, that's for Newtonian gravity, which is an excellent approximation of gravity due to curvature of spacetime - see the theory of relativity - when you aren't going a significant fraction of lightspeed). You don't experience less gravity in a fast car than in a stopped one, for example, even though cars are much faster than bikes. How bicycles work is a whole different physics question, though. $\endgroup$
    – CBHacking
    Commented Dec 12, 2015 at 7:15
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    $\begingroup$ @CBHacking - What centrifugal force? You have created a false analogy by comparing buoyancy (which is a real force) with centrifugal force (which is a fictitious force). You do not need to invoke relativity to explain why things orbit. $\endgroup$ Commented Dec 12, 2015 at 10:49

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