What happens when two objects of different masses are dropped from the same height?

See the video Apollo astronauts do this test on lunar surface. Here's the famous footage of the Apollo 15 astronaut that dropped a hammer & feather on the moon to prove Galileo's theory that in the absence of atmosphere, objects will fall at the same rate regardless of mass.

Youn tube video.

QUESTION #6

Asked by: Terri If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. This statement follows from the law of conservation of energy and has been demonstrated experimentally by dropping a feather and a lead ball in an airless tube. When air resistance plays a role, the shape of the object becomes important. In air, a feather and a ball do not fall at the same rate. In the case of a pen and a bowling ball air resistance is small compared to the force a gravity that pulls them to the ground. Therefore, if you drop a pen and a bowling ball you could probably not tell which of the two reached the ground first unless you dropped them from a very very high tower. Answered by: Dr. Michael Ewart, Researcher at the University of Southern California The above answer is perfectly correct, but, this is a question that confuses many people, and they are hardly satisfied by us self-assured physcists' answers. There is one good explanation which makes everybody content -- which does not belong to me, but to some famous scientist but I can't remember whom (Galileo?); and I think it would be good to have it up here. (The argument has nothing to do with air resistance, it is assumed to be absent. The answer by Dr. Michael Ewart answers that part already.) The argument goes as follows: Assume we have a 10kg ball and a 1kg ball. Let us assume the 10kg ball falls faster than the 1kg ball, since it is heavier. Now, lets tie the two balls together. What will happen then? Will the combined object fall slower, since the 1kg ball will hold back the 10kg ball? Or will the combination fall faster, since it is now an 11kg object? Since both can't happen, the only possibility is that they were falling at the same rate in the first place.

Sounds extremely convincing. But, I think there is a slight fallacy in the argument. It mentions nothing about the nature of the force involved, so it looks like it should work with any kind of force! However, it is not quite true. If we lived on a world where the 'falling' was due to electrical forces, and objects had masses and permanent charges, things would be different. Things with zero charge would not fall no matter what their mass is. In fact, the falling rate would be proportional to q/m, where q is the charge and m is the mass. When you tie two objects, 1 and 2, with charges q1, q2, and m1, m2, the combined object will fall at a rate (q1+q2)/(m1+m2). Assuming q1/m1 < q2/m2, or object 2 falls faster than object one, the combined object will fall at an intermediate rate (this can be shown easily). But, there is another point. The 'weight' of an object is the force acting on it. That is just proportional to q, the charge. Since what matters for the falling rate is q/m, the weight will have no definite relation to rate of fall. In fact, you could have a zero-mass object with charge q, which will fall infinitely fast, or an infinite-mass object with charge q, which will not fall at all, but they will 'weigh' the same! So, in fact, the original argument should be reduced to the following statement, which is more accurate:

If all objects which have equal weight fall at the same rate, then _all_ objects will fall at the same rate, regardless of their weight.

In mathematical terms, this is equivalent to saying that if q1=q2 then m1=m2 or, q/m is the same for all objects, they will all fall at the same rate! All in all, this is pretty hollow an argument.

Going back to the case of gravity.. The gravitational force is

( G is a constant, called constant of gravitation, M is the mass of the attracting body (here, earth), and m1 is the 'gravitational mass' of the object.)

And newton's law of motion is

where m2 is the 'inertial mass' of the object, and a is the acceleration.

Now, solving for acceleration, we find:

Which is proportional to m1/m2, i.e. the gravitational mass divided by the inertial mass. This is our old 'q/m' from the electrical case! Now, if and only if m1/m2 is a constant for all objects, (this constant can be absorbed into G, so the question can be reduced to m1=m2 for all objects) they will all fall at the same rate. If this ratio varies, then we will have no definite relation between rate of fall, and weight.

So, all in all, we are back to square one. Which is just canceling the masses in the equations, thus showing that they must fall at the same rate. The equality of the two masses is a necessity for general relativity, and enters it naturally. Also, the two masses have been found to be equal to extremely good precision experimentally. The correct answer to the question 'why objects with different masses fall at the same rate?' is, 'beacuse the gravitational and inertial masses are equal for all objects.' Then, why does the argument sound so convincing? Since our daily experience and intuition dictates that things which weigh the same, fall at the same rate. Once we assume that, we have implicitly already assumed that the gravitational mass is equal to the inertial mass. (Wow, what things we do without noticing!). The rest of the argument follows easily and naturally...

Answered by: Yasar Safkan, Physics Ph.D. Candidate, M.I.T.

I am answering this in my own way critics are welcomed.

So, Generally consider two bodies of same shape,surface area (same) but MASSES DIFFERENT.

Now if you release both the bodies from a similar height say h, with u =0m/s for both (initially rest or aka free fall) they travel h height in same time .

So,what is happening here is we know Newton's law of gravitation like F = (GMm)/R^2

Where, G is Grav.const. and M and m are masses along which gravitation (attraction only always) acts and R is the distance between the center of masses or simply centers of body (We just take point objects usually dw bout it).

Now F=[(GMē) m]/R^2 Mē= earth mass and R dist.between earth center and object

You can simplify (GMē/R^2) as g' usually when distance of object from earth surface is very less than earth radius you can take g'=g =9.8m/s^2

So,finally F=gm=mg

For two different masses This F so called gravitational force of attraction is different but look g is same for both masses that is acceleration of two bodies is same even though there is different mag.of force on them .

Mostly ppl are confused here they generally interpret this as greater the force greater the acc.n which is not right to be honest

Greater force doesn't mean greater acc.n greater force maybe because of their masses being different but with same acc.n so it is possible for two bodies of different masses experiencing different forces having same acc.n

It is more right if you put this explanation of so called g being same for two bodies rather than a1=a2=g YOU BETTER USE a1=F1/m1 and a2=F2/m2 and then you can see that g=F1/m1=F2/m2 You will be getting a lil clarity if you do it this way .

Therefore since same acc.n it says both bodies have same change in their velocities over equal time interval so these bodies attain same final velocities over a particular time since they started with 0m/s at same time that's why they cover equal disp.during their journey down hence they reach ground in same time.