Does Mass Affect Falling Speed with No Air Resistance?

[AznBoi]

Please help, I'm confused with an "easy" concept.

Suppose you drop an object off a cliff. Does increasing the mass of the object you drop decrease the time it takes the object to reach the ground? (Assume that air friction is neglected) Is this true to infinite extent? So if there was no air friction (not possible in the real world), would a grand piano land at the same time as a pencil? I just need to understand this concept as there is no way I can test it to this extent lol. I want to believe =[

 

[OrbitalPower]

When fluid mediums such as air (and any other resistance) are theoretically removed the object is said to be in free fall, which means that they fall only under the influence of gravity. In that case if the grand piano and the pencil were dropped from the same height then at a certain time t they would be going the same speed (velocity) and would have fallen the same distance.

 

[Dick]

You know gravitational force is proportional to the mass of the falling body. You also know that acceleration is proportional to force over m. Putting these together tells you that acceleration is independent of mass. You can cancel the mass. As OrbitalPower says, that's 'free fall'.

 

[AznBoi]

So the gravitational force you are talking about is F=gm1m2/r^2?? m1 being the object and m2 being earth? F=m1a, a=F/m1 Is that why the acceleration of the object is independent of its mass? Are those equations correct?

 

[Dick]

Quite correct.

 

[AznBoi]

Ok thanks a lot to both of you! =]

 

[arildno]

It should be noted that there is a tiny, theoretical effect present that makes heavier objects impact Earth before lighter.
The Earth itself experiences a greater acceleration with respect to a heavier one, hence the relative acceleration between the object and the Earth is greater, and therefore, the heavier object will impact the Earth first.

Having two objects with masses m, M, m<M, whereas the Earth's mass is $M_{e}$, the ratio of impact times is:
$$\frac{t_{m}}{t_{M}}=\sqrt{\frac{M_{e}+{M}}{M_{e}+{m}}}$$
if we consider either one of them to be released from rest from the same distance from the Earth.

 

[christianjb]

A) This is one of the profoundest questions in the history of science- so don't be ashamed in asking it. It involves the equivalence of inertial and gravitational mass, which is certainly not obvious.

B) Yes, you can test these ideas. I'd advise you though to do it safely- don't be dropping objects of tall buildings without thinking things through!