Mass doesn not effect speed of dropped objects

D H

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

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

force, mass, and acceleration: F = ma

russ_watters

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 figured you did - I was confirming and amplifying.

mikelepore

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