In circular motion, with a constant distance, why does the mass of the orbitting object have no effect on its revolution at all? 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? 
 A: 
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.
A: 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 $g$.
A: 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.
A: 
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).
A: 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.
