Mass doesn not effect speed of dropped objects

[Doc Al]

What I meant by feel was be influenced by, as in it does not make sense to say that an object in free fall towards the Earth feels a combination of the Earth's gravity and its own because it never accelerates faster than 9.8 m/s²

What do you mean by "feels a combination"? The gravitational force that Earth and object exert on each other is given by Newton's law of gravity. The resulting acceleration is given by Newton's 2nd law. Basic stuff.

yet when it hits the surface of the Earth it does feel (is influenced by) both forces.

The additional forces exerted on the object upon collision with the Earth have nothing to do with gravity.

 

[ernestpworrel]

What do you mean "resulting acceleration" Doc? Are you talking about the net force between the Earth and another planet? What does that have to do with the force imparted by the Earth to the planet? I just thought of a perfect analogy of what I'm getting here. A pair of apples is not always a pair of apples because sometimes an orange is added to it giving you a total of three pieces of fruit. Now I know what you will use as an example to illustrate your point to me so don't bother. It would be nonsense.

 

[ernestpworrel]

Oh, by the way no one has told me what they think f=ma means. Smugly referring me to Wikipedia does no good.

 

[Doc Al]

What do you mean "resulting acceleration" Doc?

I'm talking about the acceleration of a falling object due to the gravitational force between it and the earth.

Are you talking about the net force between the Earth and another planet?

Huh? Why are you dragging in another planet?

What does that have to do with the force imparted by the Earth to the planet?

Huh? Review Newton's 3rd law.

I just thought of a perfect analogy of what I'm getting here. A pair of apples is not always a pair of apples because sometimes an orange is added to it giving you a total of three pieces of fruit. Now I know what you will use as an example to illustrate your point to me so don't bother. It would be nonsense.

Huh? More nonsense.

Oh, by the way no one has told me what they think f=ma means.

What you mean what "they think"? Newton's 2nd law is basic stuff. Why not open a physics book and learn about it? In a nutshell, it relates the net force on an object to its mass and acceleration. For an object in free fall, the only force (and thus the net force) is the Earth's gravitational pull on it.

Why don't you tell us what you think "F = ma" means?

 

[D H]

What I meant by feel was be influenced by, as in it does not make sense to say that an object in free fall towards the Earth feels a combination of the Earth's gravity and its own because it never accelerates faster than 9.8 m/s² yet when it hits the surface of the Earth it does feel (is influenced by) both forces. I'm curious to know what you think I meant.

I didn't have any idea what you meant. That is why I asked.

We observe everything released near the surface of the Earth to accelerate Earthward at 9.81 m/s² because everything we release has a mass that is negligible compared to that of the Earth. To see an object accelerate Earthward at 9.82 m/s², the released object would have to be incredibly massive: about 1/12 of the Moon's mass, or an iron sphere about 1140 km in diameter.

Suppose we did just that: Construct a 1140 km diameter iron sphere, raise it a short distance above the Earth, and let it fall. From the perspective of an inertial observer, the sphere would accelerate Earthward at 9.81 m/s² while the Earth would accelerate toward the sphere at 0.01 m/s².

We can't do that, but nature does do exactly that for us. The ratio of the Earth's mass to that of the Moon is about 81:1. The Earth and the Moon are orbiting each other. (The Earth-Moon barycenter is displaced from the center of the Earth by about 2/3 Earth radii.) Suppose the Moon's mass was negligible compared to that of the Earth. This would make a month be about four hours longer than it actually is.

Ernest, you appear to have a deep misunderstanding of Newton's laws of motion and of Newton's law of gravitation. I gave you some reading material. There is plenty more available on the web and in your local public library. Please take advantage of these resources.

 

[ernestpworrel]

Why don't you tell us what you think "F = ma" means?

Here is what I think. A mass m that is has acceleration a will impart a force F to any object it collides with. In other words, force F of impact is equal to the mass m times the rate of acceleration a.

 

[russ_watters]

That's worded very badly: the first sentence isn't true the way it is worded, but the second basically is. Regardless, F=ma isn't just for impacts, it is for all accelerations - so I think you are confusing yourself by trying to consider collisions.

Anyway, there is no need for all of this running around in circles and trying to figure out what people mean by their usages of certain words. It isn't going anywhere or achieving anything. Newton's equations for gravitational force and for acceleration are listed in post #2 in this thread. If you can't figure out how they work just by looking at them, you'll need to actually start using them.

Here's what I want you to do (and this isn't optional: this thread is not going to continue going around in circles):

1. Use the gravitational force equation to calculate the gravitational force between the Earth and a 1 kg object and a 2 kg object at its surface to 4 decimal places.

2. Use the force, mass and acceleration equation to calculate the acceleration of each object under the force you calculated in step 1 (also to 4 decimal places).

3. Compare the results and tell us what you have learned.

Btw:
The mass of the Earth is 5.9736 x 10^24 kg
G = 6.6730 x 10^-11 m^3/kg-s^2
You now have everything you need to do this problem available in the thread.

 

[ernestpworrel]

What! What force mass? What is that?

 

[Jimmy]

Use the force mass and acceleration equation...

force, mass, and acceleration: F = ma

 

[russ_watters]

force, mass, and acceleration: F = ma

Sorry, yes - I missed a comma in there. It's corrected now.

The equation relates the three and can be used (as can any 3 term equation, typically) to find whichever of the three you don't know, given the other two. Obviously, it needs to be rearranged in order to be used to calculate acceleration...

 

[Jimmy]

I understood exactly what you meant, Russ. I just posted for ernest's sake. I get the impression that he's being deliberately difficult.

 

[russ_watters]

I understood exactly what you meant, Russ. I just posted for ernest's sake. I get the impression that he's being deliberately difficult.

I figured you did - I was confirming and amplifying.

 

[mikelepore]

The way I like to say that is:
F=ma
a=F/m
The ratio (F_big / m_big) is equal to the ratio (F_small / m_small).
Both are the same value of "a".

What is F_big and F_small?

I meant, for example, drop two objects near the surface of the earth.

Drop a 10 kg object (the smaller mass) and a 100,000 kg object (the bigger mass).

For each object, express its acceleration by expressing the ratio of the gravitational force that the Earth exerts on the object to the mass of the object.

10 kg object:
a = F_small / m_small = 98 N / 10 kg = 9.8 m/s^2

100,000 kg object:
a = F_big / m_big = 980,000 N / 100,000 kg = 9.8 m/s^2

The two objects experience the same acceleration.

 

[TurtleMeister]

I meant, for example, drop two objects near the surface of the earth.

I understand. Thanks for the clarification. I was thinking of the force between the Earth and only one object. My fault in misunderstanding.