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Why don't objects with greater mass fall faster?

Galileo found in his experiment that all objects regardless of its mass take same time to fall from above, if the air resistance is effectively overcome.

Aren't objects with larger mass supposed to fall quickly, since objects with larger mass attract the earth more strongly than smaller ones, and so there is stronger gravitational force between earth and objects with larger mass than between earth and objects with smaller mass?

I must be missing something.
Reply 1
Original post by MedQ

I must be missing something.


F = ma

Not F = a
Newton's Law of Universal Gravitation says the force due to gravity between two bodies is proportional to the sum of the masses of the bodies divided by the square of the distance between them.

In practical terms, the gravity force acting between either of any two bodies on Earth's surface and Earth is similar since the mass of Earth totally dwarfs the mass of any body on it.
Original post by MedQ
x

Yes, objects with larger mass experience a stronger gravitational pull. However, a larger force is needed to move them, so this balances out.

Picture this: you and your friend jump out of a plane and begin to fall. If you both hold hands mid-fall do you think you would both suddenly start accelerating twice as quickly?
Reply 4
If all Gravitational energy converts to Kinetic energy..

GPE = mgh
KE = 0.5mv^2

If they convert then mgh=0.5mv^2
Mass cancels on both sides so mass is irrelevant
Original post by MedQ
Galileo found in his experiment that all objects regardless of its mass take same time to fall from above, if the air resistance is effectively overcome.

Aren't objects with larger mass supposed to fall quickly, since objects with larger mass attract the earth more strongly than smaller ones, and so there is stronger gravitational force between earth and objects with larger mass than between earth and objects with smaller mass?

I must be missing something.


I think you have an interesting point. Obviously if you drop two masses at the same time, they both attract the Earth upwards, so the earlier replies are right. But if you drop say a 1 kg mass and then a 2kg mass and measure their acceleration towards the Earth, I reckon you would ( in theory) get different answers. But not very different, because the acceleration of the Earth is extremely small in both cases. If you work this all out, it's probably best to choose as your frame of reference the centre of mass of the Earth and the dropped object.
Original post by MedQ
Galileo found in his experiment that all objects regardless of its mass take same time to fall from above, if the air resistance is effectively overcome.

Aren't objects with larger mass supposed to fall quickly, since objects with larger mass attract the earth more strongly than smaller ones, and so there is stronger gravitational force between earth and objects with larger mass than between earth and objects with smaller mass?

I must be missing something.


It's true that a larger mass experiences a larger force due to the Earth's attraction.
However, a larger mass needs a larger force to accelerate it. The fact is, this cancels out and all masses accelerated with the same acceleration.
The maths is here

The force acting down on a mass m is, by the law of gravitation

F=GMmr2F= G \frac{Mm}{r^2}
where M is the mass of the Earth and r the distance between the mass and the Earth's centre

We know F=ma so the acceleration of this mass will be
a=Fm a =\frac{F}{m}
so subbing for F from the 1st equation you get

a=(GMmr2)m a = \frac{(\frac{GMm}{r^2})}{m}

the m cancels and you get that the mass accelerates at
a=GMr2 a = \frac{GM}{r^2}

which does not depend on m, just on the gravitational constant G, the mass of the Earth and the distance r.

(Neglecting air resistance)
(edited 10 years ago)
Original post by MedQ
Galileo found in his experiment that all objects regardless of its mass take same time to fall from above, if the air resistance is effectively overcome.

Aren't objects with larger mass supposed to fall quickly, since objects with larger mass attract the earth more strongly than smaller ones, and so there is stronger gravitational force between earth and objects with larger mass than between earth and objects with smaller mass?

I must be missing something.


It is counter intuitive because, in Earth, we experience things that way - heavier objects seem to fall faster than lighter objects. But if you go to space, say moon, where there is no significant air resistance and gravity is the only acting force on the object, then if you put two objects at the same height, one with 1000 tonnes mass and the other a 1 kilogram object, and release them, they will accelerate with the same speed and hit the surface at the moon at the same time.

Now, sometimes, that is not the case in Earth because of our atmosphere causing air resistance, and this the equation is not just F=ma=mg where a=g, anymore, but F = Mg - (air resistance), = mg - 0.5*density*velocity^2*drag_coefficient*surface_area = Ma. This is why a feather falls slower than a bowling ball in Earth, but put them in the moon and they will fall at the same time from the same height.


Posted from TSR Mobile
(edited 10 years ago)
Original post by Good bloke
Newton's Law of Universal Gravitation says the force due to gravity between two bodies is proportional to the sum of the masses of the bodies divided by the square of the distance between them.

In practical terms, the gravity force acting between either of any two bodies on Earth's surface and Earth is similar since the mass of Earth totally dwarfs the mass of any body on it.


This is simply not true.

a) The force does not depend on the "sum of the masses" it depends on the product
b) the force is not "similar". For example, the force acting on a 2kg mass, due to the Earth's attraction, is double that on a 1kg mass.
The reason they both accelerate equally is that the 2kg mass needs a force that is twice as great as the 1kg mass to accelerate the same amount because F=ma
(edited 10 years ago)
Reply 9
Looking at it from the inverse square law perspective, yes, a more massive object will have a greater attractive force towards the earth. However, most everyday objects falling towards the earth really are tiny compared to that of the Earth itself, which is why the difference in attractive force is negligible.
(edited 10 years ago)
Reply 10
There are a few assumptions. The objects have to be similar in shape so that air resistance doesn't have a significant effect on one and not the other.
Original post by ian.slater
I think you have an interesting point. Obviously if you drop two masses at the same time, they both attract the Earth upwards, so the earlier replies are right. But if you drop say a 1 kg mass and then a 2kg mass and measure their acceleration towards the Earth, I reckon you would ( in theory) get different answers. But not very different, because the acceleration of the Earth is extremely small in both cases. If you work this all out, it's probably best to choose as your frame of reference the centre of mass of the Earth and the dropped object.


You would get different answers in their times of free fall:

http://en.wikipedia.org/wiki/Free_fall#Inverse-square_law_gravitational_field
Original post by The Polymath
Yes, objects with larger mass experience a stronger gravitational pull. However, a larger force is needed to move them, so this balances out.

Picture this: you and your friend jump out of a plane and begin to fall. If you both hold hands mid-fall do you think you would both suddenly start accelerating twice as quickly?

thats a lovely example :smile:
Reply 13
Original post by The Polymath
Yes, objects with larger mass experience a stronger gravitational pull. However, a larger force is needed to move them, so this balances out.

Picture this: you and your friend jump out of a plane and begin to fall. If you both hold hands mid-fall do you think you would both suddenly start accelerating twice as quickly?


Thank you very much. That was a fantastic metaphor.
Reply 14
Original post by Maccman
If all Gravitational energy converts to Kinetic energy..

GPE = mgh
KE = 0.5mv^2

If they convert then mgh=0.5mv^2
Mass cancels on both sides so mass is irrelevant


Thank you
Reply 15
Original post by Alpha-Omega
It is counter intuitive because, in Earth, we experience things that way - heavier objects seem to fall faster than lighter objects. But if you go to space, say moon, where there is no significant air resistance and gravity is the only acting force on the object, then if you put two objects at the same height, one with 1000 tonnes mass and the other a 1 kilogram object, and release them, they will accelerate with the same speed and hit the surface at the moon at the same time.

Now, sometimes, that is not the case in Earth because of our atmosphere causing air resistance, and this the equation is not just F=ma=mg where a=g, anymore, but F = Mg - (air resistance), = mg - 0.5*density*velocity^2*drag_coefficient*surface_area = Ma. This is why a feather falls slower than a bowling ball in Earth, but put them in the moon and they will fall at the same time from the same height.


Posted from TSR Mobile


Thank you
Reply 16
Original post by Stonebridge
It's true that a larger mass experiences a larger force due to the Earth's attraction.
However, a larger mass needs a larger force to accelerate it. The fact is, this cancels out and all masses accelerated with the same acceleration.
The maths is here

The force acting down on a mass m is, by the law of gravitation

F=GMmr2F= G \frac{Mm}{r^2}
where M is the mass of the Earth and r the distance between the mass and the Earth's centre

We know F=ma so the acceleration of this mass will be
a=Fm a =\frac{F}{m}
so subbing for F from the 1st equation you get

a=(GMmr2)m a = \frac{(\frac{GMm}{r^2})}{m}

the m cancels and you get that the mass accelerates at
a=GMr2 a = \frac{GM}{r^2}

which does not depend on m, just on the gravitational constant G, the mass of the Earth and the distance r.

(Neglecting air resistance)


Thank you very much.

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