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Mass vs. Gravity and what our teachers have told us! 
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#1
Oct1204, 08:58 PM

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It seems that I have uncovered a great contradiction in modern thoughts on gravity. First, 2 statements that teachers have often used:
1) Gravity is dependent on the mass of an object. The higher the mass the stronger the gravitation field. 2) Two objects will fall to the earth with the same acceleration no matter what their mass. These two statement seem to contradict each other. The only possible resolution that i have come up with is that statement 2 is not actually correct. Instead there are slight differences in acceleration but these are not noticeable due to the infinitesimal mass of say a bowling ball and a feather compared to the earth as a whole. Does this mean that if there were another "earth" that had approximately the same mass and was dropped at the same time as the feather and the bowling ball, shouldn't the other "earth" fall at a much greater rate due to the immensely stronger gravitational field as opposed to the bowling ball or feather? I may be totaly out of it here but if anyone can shed some light onto this subject, I believe that it will make a most interesting conversation. 


#2
Oct1204, 09:17 PM

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The difference is how the various objects affect the earth, not the other way around.
In other words, you can't notice the earth falling towards the feather while the feather falls to earth. However, you will notice the first earth falling towards the second earth while the second earth falls towards the first! 


#3
Oct1204, 09:22 PM

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I see what you're saying, but there's an equation that answers your problem dealing with two "earths".
Force of gravity = gravity constant[(mass of object1 * mass of object2) / (distance between center of masses)^2] I've forgotten the gravity constant, but a quick serach should be able to locate it. 


#4
Oct1204, 11:07 PM

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Mass vs. Gravity and what our teachers have told us!
Then we can write the equation for the gravitational attraction between m1 and the earth: F1 = Gm1Mearth/r^2 By Newton's Second Law: F1 = m1a Substitute this in for F1 in the gravitational equation: m1a = Gm1Mearth/r^2 Notice we can divide m1 out of both sides leaving: a = GMearth/r^2 In other words the acceleration (or the rate at which a body falls) is independent of its mass. 


#5
Oct1304, 02:01 AM

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#2 Your equation, F = ma = GmM/r^2, only holds if there are two masses in the system. When you introduce a third mass you can no longer claim that the Net Force on m is equal to the force of gravity. The equation you used presupposes that the ONLY force on the object is the force of gravity caused by the mass M The original poster is right, anyway. This statment: Certainly Mars doesn't fall to the earth with the same acceleration as a bowling ball? Of course not, other forces interfere. In the same vein, dropping a bowling ball and a feather truly AREN'T isolated with the Earth. They each have an effect on each other and the gravity of the sun and the moon effects them and so does every other star in the universe. But these interferences are so small that it is a good general rule that stuff falls towards the earth at the same acceleration. But it doesn't even always hold, in our daily earthbound lives. Tides for example. They defy the Earth's gravity and don't accelerate towards the Earth at the same rate as a bowling ball. This is because the moon inteferes of course. Tides are proof that statement 2 is only correct under certain circumstances. 


#6
Oct1304, 05:59 AM

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[tex]GravitationalForce = \frac{GravitationalConstant*MassA*MassB}{DistanceBetweenMasses^2}[/tex] but also, the higher a mass of an object the less a fixed force will acccelerate it according to: [tex]GravitationalForce = MassObject*Acceleration[/tex] or [tex]Acceleration= \frac{GravitationalForce }{MassObject}[/tex] so on earth, [tex]GravitationalForceBowlingBall = \frac{GravitationalConstant * MassBowlingBall*MassEarth}{RadiusEarth^2}[/tex] and [tex]AccelerationBowlingBall = \frac{GravittionalForceBowlingBall}{MassBowlingBall}[/tex] will result in: [tex]AccelerationBowlingBall = \frac{GravitationalConstant * MassBowlingBall*MassEarth}{RadiusEarth^2*MassBowlingBall}[/tex] which means: [tex]AccelerationBowlingBall = \frac{GravitationalConstant*MassEarth}{RadiusEarth^2}[/tex] Thus, as you can see although the force does depend on the mass, the acceleration is independent of the mass, because the higher the mass the less it accelerates if a certain force acts on it. 


#7
Oct1304, 06:52 AM

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The weight W is defined as W = mg. Thus, already you can see it is dependent on the mass in question. However, the gravitational acceleration of that mass is g. It is independent of how large this mass is (at least, within this simple scenario where most terrestrial mechanics are applied). So next time, when you use the word "gravity", try to think what exactly it is that you are using it as. You will learn that in physics, "words" and phrases are ambiguous and vague. It is only when we realize the underlying mathematical description of it does it now become clear and unambiguous. So there are no contradictions. Zz. 


#8
Oct1304, 08:56 AM

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The first statement is that F=GmM/r^2. Combine this with Newton's second law, F=ma, and you get the equation ma=GmM/r^2. If you just divide both sides by m, this simplifies to a=GM/r^2. So, the acceleration is clearly independent of the mass m. As you can see, the second statement does not only not contradict the first. It is a consequence of the first. (Assuming the validity of Newton's second law, and the equivalence between inertial and gravitational mass). 


#9
Oct1304, 09:23 AM

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There is one thing that has (I believe) been overlooked in all the prior posts (it's a long thread, with a lot of good stuff in it, so I apologize if it was already covered):
We know that the earth also accelerates (theoreticaly) toward the other object, it is just so small an acceleration that it cannot possibly be measured. Gravitational acceleration always turns out to be equivalent to the gravitational field strength that the object is placed in. Therefore the earth will have a greater acceleration toward the more massive object than it will toward the less massive object. There would then be an a greater net acceleration of the more massive object from the earth's frame of reference. Again, this extra amount is purely theoretical and could not possibly be detected. My pet peeve: I wish teachers and textbooks would stop calling g the "acceleration due to gravity." It makes more sense to call it from the start "gravitational field strength" using the unit N/kg. FOrce of gravity = mg whether it is accelerating or not. N's 2nd law says a = F/m . When F = weight, then a = mg/m = g. 


#10
Oct1304, 09:28 AM

P: 395

Wouldn't "dropping" a huge mass on the earth result in faster acceleration though? I thought the two masses would accelerate toward each other. (In this case the moon accelerating toward the earth at the earth's acceleration of gravity rate add to that the earth accelerating toward the moon at the moon's acceleration of gravity.)



#11
Oct1304, 09:28 AM

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"It is independent of how large this mass is (at least, within this simple scenario where most terrestrial mechanics are applied)." Considering that this person is already confused between the simple concept of "gravitational acceleration" and "gravitational force", I didn't want to invoke a more complicated discussion of such things. So no, I didn't "overlook" such possibility. I just didn't want to complicate things further. Zz. 


#12
Oct1304, 12:31 PM

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Oh, so you did.
I was rushing and looking for key words. This is a discussion that comes up every year in my AP class, and I thought it was worth mentioning specifically here. 


#13
Oct1304, 12:33 PM

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#14
Oct1504, 11:06 AM

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#15
Oct1504, 11:57 AM

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But, we can turn the equation around and look at it from the Earth's point of view. We will find that if we substitute Mearth in to the F = ma part of the equation we will be able to divide out the mass of the earth and find that all objects will fall toward the bowling ball at an acceleration independent of their mass, but much much smaller than the bowling ball will fall towards the earth because the bowling is so much less massive than the earth. So, the reason for the different accelerations is that the masses are different. The earth attracts the bowling ball with a force proportional to its mass and the bowling ball attracts the earth with a force prorportional to its mass which is many orders of magnitude less than the earth's mass. 


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Oct1504, 12:04 PM

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#18
Oct1504, 02:32 PM

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The distance between the centres is D. Then the distance of the bigger body from the CM is smaller than that of the smaller body. If the masses are M and m and the distances R and r, respectively, r=D*M/(M+m) and R=D*m/(m+M). As the distance between the bodies decreases, the bodies accelerate at different rates as they fall towards each other. [tex]\frac{d^2r}{dt^2}=\frac{M}{M+m} \frac{d^2D}{dt^2}\mbox{ and } \frac{d^2R}{dt^2}=\frac{m}{M+m}\frac{d^2D}{dt^2}[/tex]. With a little math, we can write the second time derivative of D. For the smaller body: [tex] m\frac{d^2r}{dt^2}=\gamma \frac{mM}{D^2}[/tex] Replacing r with D*M/(m+M), [tex] \frac{mM}{M+m} \frac{d^2D}{dt^2}=\gamma \frac{mM}{D^2}[/tex] Rearranging a little: [tex]\frac{mM}{M+m}\frac{d^2D}{dt^2}=\gamma\frac{\frac{mM}{M+m}*(m+M)}{D^2}[/tex] [tex] \mu=\frac{mM}{M+m} [/tex] is called "the reduced mass" of m and M. We get an equation of motion which looks like that for a single body of mass [tex]\mu[/tex] in the field of a central mass of (m+M). Its acceleration is proportional to the sum of the two masses. In this way, the acceleration of both a feather and the Earth towards their common CM is lower than those of a big stone and Earth  in principle. ehild 


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