Since Newtons laws on gravity state that the gravitational attraction between two bodies is directly proportional to their masses and inversely proportional to the square of the distance between them, should a heavy object not have more gravitational force as compared to a lighter one, and consequently, the heavier object should fall faster on earth, as compared to the lighter one, when dropped from the same point, at the same time?
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$\begingroup$ Consider the relation $a=\frac{F}{m}$. Suppose you make the mass $k$ times m. Then since the gravitational force is increase $k$ times, then $a=\frac{kF}{km}$ giving the same acceleration. $\endgroup$– ProscionexiumCommented May 30 at 11:23
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/3534/2451 and links therein. $\endgroup$– Qmechanic ♦Commented May 30 at 11:35
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2 Answers
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The force on the heavier object is larger but it also has more inertia (resistance to motion). The same quantity, m, is used for mass in calculating the force of gravity, F, and in determining the object's motion using Newton's Second Law, a = F/m, and cancels out, making the motion independent of mass.
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Newton's second law is $F=ma$, where the force for the gravitational attraction in the surface of the earth is approximately $F=mg$. This means that $$ ma = F = mg \implies \mathbf{a = g}. $$ Thus we see that the acceleration of any object caused by a gravitational field is independent of its mass. We can play the same game considering $$ F = \dfrac{GmM_{\text{earth}}}{r^2} \implies a = \dfrac{GM_{\text{earth}}}{r^2}, $$ to obtain the exact same conclusion.
Thus two objects will take the same time to fall, independently of their mass. Be careful if you try this at home, because you have to also consider the drag force of the air, which will slow more the lighter object.
If you try to drop a feather and a hammer at the same time, the hammer will reach the ground first because of this drag. But if you do this in the moon, where there is no air, you see that they fall at the same time! On earth you can obtain a similar result if you put the feather and the hammer inside two equal shaped boxes (with a negligible mass).
answered May 30 at 11:28
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$\begingroup$ As a side note, the realization that the acceleration is independent of the mass, lead to the equivalence principle and to the theory of relativity! $\endgroup$ Commented May 30 at 11:29
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$\begingroup$ If gravitational force is given by F=mg, then by increasing m to M (say 2 times m), will the force not increase (F=[2m]*g)? and with higher pull, the heavier ball should fall faster... Although I know this is proved to be incorrect, I am trying to understand the falacy in my understanding. or maybe there is some other explanation. $\endgroup$– NiranjanCommented May 30 at 11:36
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$\begingroup$ Yes, the problem is that the acceleration is $a=F/m = mg/m = g$, so if you now have a mass $2m$, the acceleration will be $a'=F'/2m = 2mg/2m = g$. $\endgroup$ Commented May 30 at 12:49