Falling objects: help an ignorant fool

[russ_watters]

Uh, well, basically you misunderstood everything that was discussed in the thread and then added more misunderstandings to that. Essentially everything in that post is wrong. I really don't feel like going through the whole thing again right now, but maybe later. For a start, you can try rereading the thread a little more carefully...

Anyway, haven't you ever seen the lab experiment with a feather and a rock in an evacuated glass cylinder? Saying that small objects fall to the Earth at the same rate is correct to within a few billionths of a percent for virtually all scenarios anyone ever sees. You've allowed a mostly irrelevant caveat to confuse you.

 

[russ_watters]

Ok...

You are right about taking the bowling ball to the moon and to the earth. The Earth will fall faster, proving that heavier bodies fall faster.

Reread those two sentences. Where's the heavier body if you are asking only about the acceleration of the earth? The Earth will fall faster towards a heavier object than a lighter one, but a heavier object will not fall towards Earth faster than a lighter one (as seen from a stationary reference frame). All you've done is confuse yourself here...

You are exactly right. And Artistotle was right also.

No and no. Aristotle believed that acceleration due to gravity of an object toward Earth (using Earth as the reference frame) was directly proportional to the mass of the object. That isn't even true in your horribly twisted scenario.

This is because "g" is not a constant. It is only a constant at the surface of the earth. But "g" can be measured at any altitude and it will be different.

No. The variation of g with altitude is irrelevant here because the problem involves two objects that are both at the same distance from earth.

But the surface of the Earth is the only place it can be measured because the scientific apparatus that is used to measure "g" has to sit on something; it can't be floating in space. Remember, "g" is an experimental (empirical) value and is arrived at by experiment.

No. You can calculate it from the speed of an orbiting spacecraft , or measure it from a plane, among other things.

You cannot assume it equals (a) from F = ma. (a) is an inertial (absolute) value. But (g) is a relative value just as you said.

If it didn't, f=ma would be wrong... What you are missing is that g changes with altitude because F changes with altitude. And while I already said this is irrelevant for this discussion, if you want to play with it, you can easily use Newton's gravitation equation to calculate variations in g with altitude by plugging it into f=ma.

If the apparatus is sitting on the Earth that is moving toward the bowling ball, it is measuring the acceleration of the Earth toward the bowling bowl plus the acceleration of the bowling ball toward the earth.

Yes! However, I don't think there is a device anywhere on Earth sensitive enough to differentiate those two accelerations.

Therefore g = a + a' and with a little algebra you can show that g = G(m + m')/R^2.

Not sure about what you did with that equation, but the way I would do it is to simply calculate two different accelrations with f=ma. Newton's gravity equation gives you one force and you have two different masses.

This is why all bodies fall at the same rate; because as you pick up a rock, the Earth gets smaller but the quantity (m + m') stays constant, therefore "g" is constant.

That has nothing to do with anything. This discussion works fine if you take a spaceship and fly it to another planet.

All objects that originate from the Earth fall back at the same rate. This is why Galileo was confused. But if the object comes from outer space (holding the Earth constant) the heavier object will fall faster. This is what you already proved with the bowling ball and the earth/moon example.

Whoa, back up a sec. Above, you said Aristotle was right. Aristotle said that the acceleration due to gravity was directly proportional to mass for small objects on earth. That is most certainly not true.

All you are really proving by bringing up the idea of picking up a rock is that the caveat we've been discussing here is useless. As I said before, for small objects like rocks and bowling balls and feathers, the difference in acceleration due to gravity if you include the Earth's acceleration or not is so small that it won't even fit in the decimal places available in your calculator and should be ignored for that reason. Only when the smaller object is perhaps .1% of the mass of the larger one or larger, does that effect become even measurable, much less important enough to actually calculate.

 

[Smoke/mirrors]

Hi, Watters. Scifell made some very profound observations in his first post and I can't help but agree with him. He may think he is an ignorant fool, but I, for one, don't think so. I think he is more savvy than all the people that argued against him in this thread. He understands relative acceleration and has a keen perspective on physical processes. With a little math background, he is bound to go places. He was right in questioning all three of his observations. I will stick to what I said. "All bodies that originate from the Earth will fall back to the Earth at the same rate. All bodies that originate from outer space will fall faster." The equation I developed will prove that when you plug in the numbers for the Earth (m') and the bowling ball (m). It also works for the moon and the bowling ball. It works for any two bodies in space. Here it is again

*** g = G(m + m')/R^2 ***

Try it. Aristotle was RIGHT. Galileo was only HALF RIGHT. And Scifell wass RIGHT TOO!