Two object run together to the earth despite of their mass

Hepic · Sep 7, 2013

Everybody knows that two object run together to the earth despite of their mass. For example we have a object with 30 kg and one with 50 kg and let them from a big height. The reason that they will reach at same time will be that the earth will put different power about their mass so to catch the asseleration of 10m/s^2. Is that right?

christianpoved · Sep 7, 2013

Hey! the answer is "no", that's not right this doesn't have anything to do with power. Everything falls with the same acceleration because this acceleration is independant of the mass, some easy calculations can be done to show it.

Drakkith · Sep 7, 2013

Remember that power is work per unit of time, such as joules per second, and is expressed in Watts.
And work is force acting over a distance. W=fd.
So you are correct in that the larger object has "more power", as it can perform more work than a less massive object, but we don't usually put it that way as it isn't really correct. Instead, we say that a greater force is applied to the more massive object. This increase in force is exactly counterbalanced by the increase in mass, thus leading to the acceleration being exactly the same between objects of different masses.

Bandersnatch · Sep 7, 2013

I think Hepic has got the right idea, only uses non-physical nomenclature.

What he says is that the force(not power) of gravity is proportional to the mass of the falling body, so in the end when we put [itex] F=F_g → ma=\frac{GMm}{R^2}[/itex] the mass m cancels out and we get acceleration independent of it:
[tex]a=\frac{GM}{R^2}[/tex]

namanjain · Sep 8, 2013

christianpoved said: ↑

Hey! the answer is "no", that's not right this doesn't have anything to do with power. Everything falls with the same acceleration because this acceleration is independant of the mass, some easy calculations can be done to show it.

Bandersnatch said: ↑

I think Hepic has got the right idea, only uses non-physical nomenclature.

What he says is that the force(not power) of gravity is proportional to the mass of the falling body, so in the end when we put [itex] F=F_g → ma=\frac{GMm}{R^2}[/itex] the mass m cancels out and we get acceleration independent of it:
[tex]a=\frac{GM}{R^2}[/tex]

exactly
why to go to that power where we have this
acceleration same then time to cover same distance is same

Buckleymanor · Sep 8, 2013

Drakkith said: ↑

Remember that power is work per unit of time, such as joules per second, and is expressed in Watts.
And work is force acting over a distance. W=fd.
So you are correct in that the larger object has "more power", as it can perform more work than a less massive object, but we don't usually put it that way as it isn't really correct. Instead, we say that a greater force is applied to the more massive object. This increase in force is exactly counterbalanced by the increase in mass, thus leading to the acceleration being exactly the same between objects of different masses.

Is this allways true for different masses of any size.A weight does not fall at the same rate on the Moon as on Earth.

paulfr · Sep 8, 2013

A non rigorous way to explain why the 50Kg behaves the same as
a 30 Kg object is the following.
The 50Kg object experiences more FORCE [F = ma = mg],
But it also has more inertia and thus more resistance to a change in its velocity.
The two effects balance to give either object the same acceleration.

jtbell · Sep 8, 2013

Note that we have an FAQ about this, in the "Frequently Asked Questions" section at the top of this forum.

https://www.physicsforums.com/showthread.php?t=511172 [Broken]

Hepic · Sep 8, 2013

Thank you.you were clear,but I have a question too.How a planet understands that an object has more mass from an other so put more force?

davenn · Sep 8, 2013

hi Hepic

so you really didn't quite understand ?

did you read the link jtbell pointed you to ?

the earth doesn't know anything ... It doesn't have a mind, to quote a comment from his link ...

What I'm trying to show here is in this scenario, the earth does not make any significant motion due to the force exerted by m 1 . This means that for this case, it is perfectly fine to consider the earth as the fixed object, and only consider that it is the smaller mass that falls towards the earth. So when you consider that, then all the simplification that is done to allow us to deduce that the acceleration due to gravity of ALL objects falling on the surface of the earth to be a constant, independent of the mass of the object.

Note the bolded section

Dave

Buckleymanor · Sep 9, 2013

What is not quite so clear is the last part of the link.

However, if we change that situation, i.e. if m 1 is now comparable, or even bigger, than M , then those simplifications will no longer be valid, and one will have to start from the very beginning to figure this out.

Small masses have a small effect on the Earth.Once the masses get quite large they do and the Earth no longer remains a fixed object.This only happens when the object is comparable in size to the Earth.The Moon is quite a bit smaller but comparable in size and would have an effect.
How large an object has to be to be comparable is not so clear.

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Two object run together to the earth despite of their mass

Hepic

christianpoved

Drakkith

Bandersnatch

namanjain

Buckleymanor

paulfr

jtbell

Staff: Mentor

Hepic

davenn

Buckleymanor