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Using Newton's universal equation and some circular motion equation, the orbiting object's mass cancels out. But can someone please explain why this is without using pure algebra?

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    $\begingroup$ Only impure algebra is allowed. $\endgroup$ – Marc van Leeuwen Mar 31 '15 at 11:33
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    $\begingroup$ FYI - the orbiting object's mass does have an effect on the shape of the orbit. For example, if you increased the mass of the moon such that it had the same mass as the earth, the moon would no longer follow a circular path with the earth near its center. Instead, the moon and earth would both orbit around a point halfway between them. See barycentric coordinates. $\endgroup$ – mbeckish Mar 31 '15 at 15:47
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    $\begingroup$ I'm trying to imagine what the world would be like if the orbit of a heavy object (like the earth) around the sum differed greatly from the orbit of a much lighter object (like me) around the sun. The picture isn't pleasant. $\endgroup$ – Andreas Blass Mar 31 '15 at 19:30
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But can someone please explain why this is without using pure algebra?

I will try without a single formula.

In Newtonian gravity, the gravitational force on a particle is proportional to the particle's gravitational mass; the more gravitational mass, the more the gravitational force.

In Newtonian mechanics, the acceleration of a particle, for a given force, is inversely proportional to the inertial mass; the more inertial mass, the less the acceleration.

If it is the case that the gravitational mass and inertial mass are equal (so that we speak only of the particle's mass), the gravitational and inertial mass cancel and the gravitational acceleration of a particle is then dependent only on the strength of the gravity at the place the particle is.

But, in the Newtonian context, it is observationally the case that gravitational mass and inertial mass are equal.

In the particular case of circular motion, the distance from the gravitational source is constant and, thus, the (inwardly, radially directed) gravitational acceleration of the particle is constant (and independent of the particle's mass).

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  • $\begingroup$ That's exactly the answer I would have written if you hadn't gotten there before me! $\endgroup$ – Floris Mar 31 '15 at 2:24
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    $\begingroup$ Is it just me, or does this answer read a lot like a verbal description of an algebraic proof? $\endgroup$ – talrnu Mar 31 '15 at 16:40
  • $\begingroup$ Good answer but I have a small quibble; I would not characterize General Relativity as explaining why gravitational and inertial mass are observed to be the same. Rather, GA starts by postulating that they are equivalent. See en.wikipedia.org/wiki/Equivalence_principle. $\endgroup$ – Eric Lippert Mar 31 '15 at 17:04
  • $\begingroup$ @EricLippert, is that true? From the article: - "Although the equivalence principle guided the development of general relativity, it is not a founding principle of relativity but rather a simple consequence of the geometrical nature of the theory. In general relativity, objects in free-fall follow geodesics of spacetime,..." - I think I'll remove that parenthetical since it is superfluous and might be controversial. $\endgroup$ – Alfred Centauri Mar 31 '15 at 17:46
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    $\begingroup$ This answer, along with most of the rest, is not quite correct. Using that forbidden algebra (per the question), the angular velocity of an orbiting body is given by the correction to Kepler's third law derived by Newton, $\omega^2 = \frac {G(m_1+m_2)}{a^3}$. Note the dependency on the sum of the masses. This answer, along with most of the others, is only correct if $m_2 \ll m_1$ (or alternatively, if $m_1 \ll m_2$). $\endgroup$ – David Hammen Mar 31 '15 at 22:33
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You certainly know that all things fall at the same rate regardless of their mass (neglecting friction).

An orbiting body is not different from a falling body in that the only force acting on it is the gravity of the thing it orbits, so there is no reason its mass should influence its orbit.

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    $\begingroup$ Wrong, in both cases. The mass of the falling object does matter. The rate the object falls is based on the sum of the masses of the two objects, not merely on the mass of the object it's falling towards. It's just that in normal examples the masses are so different that the effect is far smaller than the experimental error. When you are dealing with objects of roughly similar mass it most certainly does matter. $\endgroup$ – Loren Pechtel Apr 1 '15 at 0:26
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Why does the mass of the orbitting object have no effect on its revolution at all?

It does have an effect! However, the effect is immeasurably small if the orbiting object itself has a very small mass compared to the object it is orbiting. The most massive object we humans have put into orbit is the International Space Station, with a mass of 419.5 metric tons (plus a bit more for any visiting vehicles that happen to be attached at the time). At less than 10-19 Earth masses, that's still quite tiny in comparison to the Earth.

The Moon on the other hand has a mass of about 0.0123 Earth masses. On magically replacing the Moon with the International Space Station, one would find the Space Station to be orbiting at a slightly reduced orbital rate compared to the Moon's orbital rate, about 0.6% slower. On magically replacing the Moon with Venus, one would find our sister planet and the Earth orbiting themselves about 34% faster than the Moon's orbital rate.

The other answers have correctly stated that the acceleration of the orbiting body toward the central body is independent of mass. What those other answers have ignored is that the central body is also accelerating gravitationally toward the orbiting body. This is insignificant if the orbiting body's mass is insignificant. It's not insignificant in the case of the Earth and the Moon, Pluto and Charon (12% of Pluto's mass), and especially Alpha Centauri A and B, whose masses are 1.1 and 0.9 solar masses.

You asked not to use algebra. The math is pretty simple. The orbital rate is a function of the sum of the masses of the two bodies. In the special case that one body's mass is many, many orders of magnitude larger than the other (e.g., the Earth and the Space Station), the sum of the masses is for all practical purposes equal to that of the larger body. In the case of Alpha Centauri A and B, you'll get a very wrong answer for the orbital rate if you don't use the sum of their masses.

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Orbit is free-fall around a body, so it is for the same reason a feather falls as fast as a bowling ball (in a vacuum).

In free-fall,

$F = mg$

And we know that,

$F = ma$

So we can substitute,

$ma = mg$

And divide by $m$,

$a = g$

Thus, no matter what mass is, acceleration equals $g$.

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    $\begingroup$ Jimmy360, to be sure, I've added an answer that makes the distinction between the gravitational mass in your first equation and the inertial mass in your second equation. In Newtonian gravity and mechanics, the two masses are distinct in principle but equal by observation. $\endgroup$ – Alfred Centauri Mar 31 '15 at 1:29
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You are no doubt familiar with the apocryphal experiment of Galileo showing that falling bodies fall at a rate independent of their "weight". [ We should really say mass.]

An orbiting body is just a special kind of falling body, albeit one that manages to miss the ground due its sideways motion. Hence the term "free fall" as used regarding astronauts or other objects in orbit.

Personally, I'm not quite clear as what else anyone might expect other than the mass cancelling. Suppose I have a two cuboidal blocks of wood 50cm long and 25cm square side in free fall. If I cut one across the middle to create two 25cm cubes, why would I expect them to start accelerating relative to the uncut block? This scenario applies whether we are dropping the blocks from a height (in a convenient vacuum filled tower), or in orbit.

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  • $\begingroup$ Beware of hindsight bias :) I'm pretty sure there were ancient Greeks who noticed the same thing - but Greek philosophy (which was also the basis for medieval philosophy, and surviving within the lay public to this day) didn't seek to find out the truth; they were just throwing hypotheses around, without trying to get rid of those that observably didn't work. Scientific approach is a very recent technique indeed, and even among scientists, it isn't always used perfectly. How could industrial-era scientists be satisfied with an explanation like "Elan Vital"? That's no explanation! :D $\endgroup$ – Luaan Mar 31 '15 at 14:03
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    $\begingroup$ I like your "cut the block in half" idea. A way I like to think of this is to imagine a small iron ball and a large iron ball, and hypothesize that the smaller falls slower. Attaching the balls together with a thin thread should cause the small ball to slow down the large ball; it drags it back. But two-balls-and-thread are themselves a heavier object than the large ball, so they should speed up. Since we conclude a contradiction, that the larger ball must be both sped up and slowed down by its connection to the smaller ball, we have reason to suspect that the hypothesis is bad. $\endgroup$ – Eric Lippert Mar 31 '15 at 17:09
  • $\begingroup$ @EricLippert. I like your ball within a ball idea too. It seems to me that as soon as you have well a defined notion of speed the result is obvious. And without having defined speed, any hypothesis about it is meaningless babble, to paraphrase the other comment. It is therefore no coincidence that Galileo is also credited with being the first to measure speed. [en.wikipedia.org/wiki/Speed] $\endgroup$ – Keith Mar 31 '15 at 23:28

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