Okay, you’re in your vacuum on Earth and drop a feather and a bowling ball. They’re going to hit the floor at the same time. Now you’re going to do the same thing on Jupiter. Why don’t they fall at the same speed?
The only answer anyone has been able to give me is that Jupiter is more massive. Yeah, so is the bowling ball, but that didn’t matter. What’s the difference?
If you’re in a vacuum chamber on Jupiter and you drop a bowling ball and a feather, they will fall at exactly the same speed.
Um, they do fall at the same speed (same as each other, that is). They don’t fall at the same speed they fell on Earth because gravity is stronger on Jupiter.
More generally, here is the equation for the gravitational force under Newton’s law (since superseded by general relativity, but good enough for this):
F = G (m1 * m2) / r^2
When you add Newton’s other big equation: F = m * a, you get a = F / m2 = G * m1 / r^2
Thus, the mass of the second item (the ball or feather) doesn’t matter - only the mass of the first (Earth or Jupiter).
To put it in less mathy terms:
The planet pulls on the ball harder than on the feather, but the ball has more resistance to being moved (being heavier) so in the end they both get moved with the same acceleration.
And a bigger planet planet pulls harder on both of them, so they both fall faster but at the same speed as each other (for the same reason as on earth).
Acceleration due to gravity will vary in proportion of the mass the object is falling toward, and inversely in proportion to the square of the distance. The mass of the object that is falling will not cause a variance (in vacuum,) as the terms cancel out.
So the speed that objects fall on the surface of the earth is dictated by the size and mass of the earth.
Note that the earth is also falling towards the bowling ball, and to an even tinier extent towards the feather. However, in practice, we can’t measure the effect, especially as there are probably falling feathers and bowling balls and other stuff on the other side of the earth too.
On Jupiter, the distances are greater, but the mass of the planet is much greater, so your bowling ball and feather fall faster. Jupiter falls more slowly towards the bowling ball and the feather, because they’re further away. But the same principles apply.
Besides the answer to the physics question, I’m curious about this impression that you had, that they’d fall at different speeds on Jupiter, and that other people you talked to seemed to agree with that impression.
First, do you recall how you got the impression that they’d fall at different speeds on Jupiter?
Second, who were these people that you asked?
Aren’t ambiguous comparisons fun?
I think what he meant was that since the bowling ball is more massive it should pull toward the earth faster than the feather because objects that are more massive have more gravity. Since the ball and feather are both pulling on the earth, why wouldnt the more massive object pull harder and therefore accelerate faster and reach the ground faster.
Jupiter was just used as an example of how more massive objects cause greater acceleration due to gravity, so why do the ball and feather accelerate equally… Thats how i read his question anyway. It isnt “why do they fall at unequal rates on jupiter?”, it’s “why do objects of different masses accelerate equally anywhere, since more mass means more gravity and more acceleration?”
And the answer is that more gravity doesn’t necessarily equal more acceleration. Newton’s law of universal gravitation gives you a force as the end result. But acceleration = force / mass (to be accelerated.) That’s where the mass of the falling body falls out of the equation when you solve for acceleration - it gets divided by itself.
But I guess the question is why doesn’t it work the other way around. Let’s say you dropped the earth and Jupiter onto a bowling ball at the same time, in a vacuum of course. According to the standard way of thinking about these things, this would mean that the earth and Jupiter would fall at the same rate. Which in turn seems to mean that that the ball would fall at the same rate on the earth and on Jupiter just by changing your perspective. But as we know, that is not what happens. So I think the question is why does the mass drop out of the smaller objects, but not the bigger ones?
Now imagine that you are dropping a bowling ball onto the earth, but you have the magical ability to increase the size of the bowling ball at will. At what point will the bowling ball begin to count as the bigger object? At exactly that point when it is the tiniest bit bigger than the earth or is something else happening here and there’s a more gradual change?
That’s what I’m taking the OP to mean. Also, I’m curious too now.
I think the answer to that question is that the arbitrarily-large bowling ball and the feather will both fall at the same speed, but the Earth will be very very slightly falling towards the bowling ball and feather at the same time. And with the bowling ball, the Earth will move up to meet the bowling ball, ever so slightly more than it will move up towards the feather, so that in theory, it will take slightly less time for the bowling ball to contact the Earth. Not because it fell faster, but because it had slightly less distance to travel.
This “slightly less” will be completely un-measurable. How far does the Earth move if you apply a force of 16 pounds for a fraction of a second?
If you work out the equations for motion in all their glory, you get a rapid motion for the bowling ball/feather moving earthward, and a very slow motion of the Earth moving featherward. For all practical purposes, you can ignore the motion of the Earth caused by the feather, because the Earth weighs so much more than the feather. When the two objects are closer in mass (dropping the Earth onto Jupiter, or Jupiter onto the Sun), you can no longer ignore the mass of the falling object.
Well, it sounds as if you’re talking about the two objects falling together, not breaking it apart into ‘the ball is falling towards the Earth at this rate, but the earth is falling towards the ball at this much smaller rate’ as I was doing in my analysis.
The answer is, there’s no point of sudden change. If you drop a 1 pound bowling ball and a 2 pound bowling ball, separately, (because if you drop them at the same time the earth is falling towards both of them at once,) and time them with sufficient precision, the 2 pound bowling ball and the earth will fall together something like 10^-25 faster than the 1 pound bowling ball did - this being the fraction by which you’ve increased the relative mass of the Earth/Ball system. So once you get to a ‘ball’ with 1% the mass of the earth, they’ll fall together approximately 1% faster, and so on.
I think that’s it, anyway - the fact that the two masses are multiplied, instead of added, in the Universal Gravitation equation is tripping me up - but now that we’re looking at the two objects falling together, then that changes the way we factor out for acceleration.
I’m pretty sure it’s a smooth transition, though, that is quantifiable but far below the threshold of measurement for all balls of reasonable size.
The earth is about 6 trillion-trillion Kg, a bowling ball about 7Kg, a feather about a gram. The difference in gravitational effect between the ball vs. the feather opposing the earth is vanishingly small, so they fall at the same rate (in a vacuum).
Jupiter is about three hundred times as massive as the earth, but it is also less than one quarter the density, so the “surface” is quite a bit further away from the gravitational focus point. Probably not far enough to make things fall more slowly, though, I guess it depends on what you call the “surface”. Gravity is somewhat lower in earth’s ionosphere, but earth does have a clearly defined hard surface which is not a certain fact for Jupiter. If you were to descend to the point where the Jovian atmosphere gives way to something else, ocean, giant emerald, what ever, I think the sheer mass of the atmosphere itself would affect the net gravitational vector and things might fall a bit more slowly.
That prompts an interesting question - would gravity increase as you descend through Jupiter’s layers? The effective mass would decrease, because any gravitational pull from the layers above you cancel out to zero, as I understand it. But you’re getting closer to the denser core.
Equations without definition of terms are less than useless.
This is a classic, solved by Newton a long time ago.
Actually, the pull of gravity decreases as you fall deeper and deeper into Jupiter, because only the mass of the sphere-shaped part of the planet below you can attract you. The mass of the big spherical shell of planet above you doesn’t have any effect either way.
(One might think that it would pull upward slightly, but Newton showed that, instead, it cancels out entirely. If you are inside a spherical shell of matter – inside the Hollow Earth, or inside a Ringworld or Dyson Sphere – the matter pulls you in all directions equally, and thus cancels out.)
OK, here:
m1 = mass of planet
m2 = mass of object falling toward planet
F = force of gravitational attraction between object and planet
a = acceleration of object toward planet
r = distance between center of object and center of planet
G = gravitational constant, 6.674x10[sup]-11[/sup] N(m/kg)[sup]2[/sup]
Did Newton solve it for the case where the density of the planet was constant, or for all possible cases? I know about the ‘shell cancelling out’ effect, and I referenced it in my earlier post. However, consider:
Take a spherical volume centered around the planet Earth, but with equal radius to the commonly accepted radius of Jupiter. Release a low-density gas into this volume, (and use a force-field if necessary to keep it from drifting away) and call this volume the planet Super-Earth.
As you ‘descend’ through Super-Earth to the solid surface, will the pull of gravity increase or decrease? I say, in this case, it will increase, because the effect of the sparse gas in the layers you are descending past is small enough to be outweighed by the fact that you are approaching closer to a dense core, and the effect of gravity from that core, (ie, the solid planet Earth as we know it,) will grow stronger as you approach.
Would you agree with me so far?
In that case, is there anything that prevents the same effect from manifesting as we descend through the Jovian atmosphere? I suspect that we don’t know density of the different layers of Jupiter to be certain.
(And yes, I’m pleased that I got an opportunity to use ‘Consider:’ after you’d referenced Ringworld 
)
chrisk is right to ask the question. For the layers that are above you, their pull cancels out so that you have only the mass remaining below you. But you are now closer to the center of that mass, and the distance term in Newton’s equation is squared, so it makes a significant difference.
I would fully expect that as you descend into Jupiter’s gas ball, the gravity would increase up to a point. I know that if you dug a tunnel into the Earth, the pull of gravity keeps increasing as you go down, until you get pretty far down. This is due to the very dense iron core at the Earth’s center.