Why does an object with smaller mass hits the ground at same time compared to object with greater mass? I understand the acceleration due to gravity of earth will be same but won't the object with greater mass will fall faster?


That is an excellent example for a nice quote I read on the internet: "Common sense may be common, but it certainly isn't sense" :-)

As it is hard to lift heavy objects, we assume that it must be easier for them to drop.

Now, Newton's laws point out that light and heavy objects will fall with the same velocity. But is there an intuitive reason? Yes!

The mass of an object contributes to two different phenomena: Gravity and inertia.

  • The heavier an object is, the stronger the gravitational pull it experiences.
  • The heavier an object is, the stronger its resistance to an accelerating force will be: Heavier objects are harder to set in motion, meaning that for the same acceleration you need a larger force.

When people think that heavy objects should fall faster, they only think of the first point. But in reality, the first and second point cancel out each other: Yes, the earth pulls stronger on a heavy object, but the heavy object is more reluctant to get moving.

  • $\begingroup$ Is it the force that acts upwards tries to get the object more reluctant to falling down? $\endgroup$
    – cpx
    Feb 27 '11 at 3:02
  • $\begingroup$ Not sure I know what you mean. The mass has two roles: First, it tells you how inert your object is via $F = ma$, which you can rearrange for $a = F/m$. So the same force acting on a heavier object results in a lower acceleration. Second, it tells you how strong gravity pulls on it. In a homogeneous field: $F = mg$ where $g$ is the gravitational constant of earth's surface. You see: $F$ is proportional to $m$ and $a$ is anti-proportional to $m$, so they cancel out each other. $\endgroup$
    – Lagerbaer
    Feb 27 '11 at 16:32
  • $\begingroup$ Btw, it's quite remarkable and all but trivial that the inert mass and the gravitational mass seem to be exactly the same. Einstein's general relativity talks in length about this. $\endgroup$
    – Lagerbaer
    Feb 27 '11 at 16:33
  • $\begingroup$ Wow.. Great answer without the use of equations!!! $\endgroup$ Dec 8 '12 at 17:04

If you understand that the acceleration is the same for both objects, then it's the same as saying that they'll fall at the same rate. Acceleration=Rate of Change of Velocity = Rate of Change of Rate of Change of Position. So, if they start at the same point and with the same initial speed and if they have the same acceleration then their velocities and positions change at the same rate and hence they'll hit the ground at the same time.

Another way of understanding this is famous thought experiment(by Galileo I think): Assume that there is a difference in the time taken to fall of two bodies with different masses. Now, take two bodies of the same mass and measure the time taken for them to hit the ground. Tie them together. This can be treated as one body of twice the mass. Then that means it will take a different time to hit the ground. But tying them together shouldn't have made any difference as it can thought of as putting the the two bodies very close to each other and then dropping them. Hence, all bodies fall at the same rate.

  • 2
    $\begingroup$ I have always loved this deductive argument for equal velocities. The way I heard it is different but I think stronger. Assume one mass is very large and the other very small. If you tie them together with a massless rope or rod, you would expect the small mass to slow the large mass down if small masses fall slower. Hence, the two masses would fall somewhere in between the two velocities at which they would fall if they were separate. However, by looking at the system as a whole, the system of two masses should fall faster than heavy mass separately since the total mass is larger. $\endgroup$
    – BeauGeste
    Feb 26 '11 at 21:11

The real reason is "inertial mass" and "gravitational mass" of an object is same in nature. In Newtonian scheme, the force with which an object is attracted towards earth is $\frac {GMm_G}{R^2}$ $M$ is the mass of the earth, $m_G$ is the gravitational mass of the object and $R$ is the distance between the center of the earth and the object. If the object has an inertial mass $m_I$ and under the force of gravity it has an acceleration $g$ then the same force is $m_Ig$.

Therefore $m_Ig = \frac {GMm_G}{R^2}$

If $m_I = m_G$ then $g = \frac {GM}{R^2}$

That is the acceleration of the object is independent of its inertial mass. Now the deeper reason of why the inertial mass and gravitation mass of an object is same is purely coincidental in Newtonian scheme, but in general relativity it is indispensable. The fundamental postulate on which the edifice of GR is based is the "equivalence principle" which requires these two kinds of masses to be the same.


The accepted answer is not generally correct. Light and heavy objects do not necessarily fall with the same acceleration. Common sense is both common and sense, except among gravitational physicists.

The rules are:
1. The inertial acceleration of a body is proportional to the mass of the attracting body, and does not depend on its own mass.
2. The relative acceleration of two bodies is proportional to the sum of their masses.
3. The time for a body to fall to the Earth is inversely proportional to the sum of the mass of the Earth and the mass of the body.

When a body is picked up to a certain height and then dropped, the time to fall to the Earth does not depend on the mass of the object. If you lift a ping-pong ball and then drop it, it will take the same amount of time to fall to the Earth as a bowling ball. Splitting the Earth into two masses does not change the sum of those masses, or the free fall time. Contrary to the other answers, the acceleration with respect to inertial space does depend on the mass of the dropped object. This is because the mass of the Earth is reduced by the amount of mass lifted, so that the total mass remains constant. Counterintuitively, a heavier body experiences less inertial acceleration than a lighter body. In this case, it is the relative acceleration that is independent of the mass of the external body.

Now, when an external body is brought to a certain height above the Earth and then dropped, the free fall time does depend on the mass of the external body, because the sum of the Earth and the body obviously depends on the mass of the body. The relative acceleration also depends on the mass of the external body. In this case, the inertial acceleration of the external body is independent of its mass, as is so often claimed.

The third scenario is when two bodies are dropped simultaneously. They will always reach the ground simultaneously. They will experience the same inertial acceleration, which does not depend on their masses.


Not the answer you're looking for? Browse other questions tagged or ask your own question.