Start with variables r = radius of Earth = 6371000 m m_e = mass of Earth = 5.972E24 kg m_gb = mass of golf ball = .05 kg m_bb = mass of bowling ball = 6 kg m_bh = mass of black hole at center of Milky Way (est) = 8.2E36 kg m_o = mass of arbitrary object F_e.gb = force between Earth and golf ball = .491 N F_e.bb = force between Earth and bowling ball = 58.92 N F_e.bh = force between Earth and the black hole = 8.052E37 N F_e.o = force between Earth and arbitrary object a_e.gb = acceleration of the Earth towards the golf ball = .00008029 m/s^2 a_gb.e = acceleration of the golf ball towards the Earth = 9.82 m/s^2 a_e.bb = acceleration of the Earth towards the bowling ball = .00963481 m/s^2 a_bb.e = acceleration of the bowling ball towards the Earth = 9.82 m/s^2 a_e.bh = acceleration of the Earth towards the black hole = 1.348E13 m/s^2 a_bh.e = acceleration of the black hole towards the Earth = 9.82 m/s^2 (corrected rounding error) a_o.e = acceleration of arbitrary object towards the Earth d = distance between objects = 10 m d_e = distance Earth falls before collision with object d_o = distance object falls before collision with Earth t_e = time Earth falls before collision with object t_o = time object falls before collision with Earth F = Gm1*m2/r^2, therefore the Force calculations above were used with the associated variables listed and can be verified here d = (a*t^2)/2 t = sqrt(2d/a) For any object 10 m above Earth, we use subscript _o: t_e = t_o = sqrt(2d_e/a_e.o) = sqrt(2d_o/a_o.e) This gives us d_e*a_o.e = d_o*a_e.o d_o = d_e*a_o.e/a_e.o We also know that d = d_e + d_o = 10 m Substituting we get d_e + d_e*a_o.e/a_e.o = d d_e = d / (1+(a_o.e/a_e.o)) This allows us to calculate t t = sqrt(2d_e/a_e.o) = sqrt(2(d/1+(a_o.e/a_e.o))/a_e.o) = sqrt(2d/(a_o.e+a_e.o)) ========================================= For the golf ball: t = sqrt(20/(9.82 + .00008029)) = 1.4271101 seconds For the bowling ball: t = sqrt(20/(9.82 + .00963481)) = 1.4264164 seconds For the black hole: t = sqrt(20/(9.82 + 1.348*10^13)) = .00000121806 seconds ========================================= The reason most people get this wrong is because it is presumed that the Earth does not also move towards the falling object. It should be obvious that this is the case when the "falling object" is extremely massive (i.e. a black hole) - QED
From another thread: RPenner, as you can see, Dr_Toad and others are requesting your mathematical acumen in the evaluation of my proof. Thanks in advance! Please Register or Log in to view the hidden image!
RJBeery, are you saying that some folks looked at this equation... F = G (m_1 * m_2) / r^2 ...and concluded that only one of the masses affects the attractive gravitational force between them?
The rate an object falls (free fall) has nothing to do with when It was dropped or the mass of the object in free-fall. Since Galileo discovered this he's technically right infinite in extent. Unless somebody can show otherwise. Which RJBerry has't done with his juvenile analy You think there is some physics which will include force and acceleration in an inertial free fall frame. RJBerry analysis reminds me of a chinglu analysis where he either adds or subtracts, something that should be or shouldn't be included in the physics. Winding up with irrelevant nonsense.
Yes, Neddy Bate, that's exactly the problem. They apparently believe that the force would only apply to one of the two masses.
If two objects are in free fall and one object falls a shorter distance before hitting the ground, I have to assume that you would agree it would do so in less time, correct? That's the crux here. The Earth is pulled up toward the bowling ball (or the moon, or the black hole) with a greater force than with the golf ball, therefore the bowling ball (et al) hit the Earth in less time because they are falling a shorter distance.
Except that while the earth is pulled up toward the bowling ball shortening the distance.., in the case of the golf ball, IT (the golf ball) falls toward the earth faster than the bowling ball does.., also shortening the distance... This is never an issue, if you always use the earth as the frame of reference for the fall in both cases.., because all of the closing velocity is then attributed to the smaller object. When you try to look at it from the rest frame before the drop you have to take into account that the golf ball falls faster toward the earth than the bowling ball does.., which is exactly balanced by the difference in how fast the earth falls toward the smaller object in each case. You are wrong! And as been pointed out Newton and Einstein are both right. Take the time to look over, The force of gravity is the same for atoms and baseballs. That should be a larger difference in mass than your bowling and base ball fantasy. And I am sure than even though I don't want to strain myself with the math, the researchers involved had no issue with the math... Without a reference I think I also remember a proposed test with neutrons, but I am unsure and a little skeptical given the short half life of a free neutron.
So are you saying the Earth accelerates toward the bowling ball, even as the bowling ball accelerates toward the Earth, and, if so, what happened to that 9.8 m/s² g-field? Because as the Earth rushes toward the bowling ball, g, which is a function of r, now needs to be treated as g(r(t)). I don't see where you covered that in your mathematical analysis. What now?
The 9.82 m/s^2 is unchanged in my analysis. The numbers I use in my analysis in the OP are DERIVED via F=Gm1*m2/r^2. They are NOT simply given by convention! The difference in g(r(t)) is indeed ignored, but that is inconsequential because the field gradation between 10 meters and the ground of the Earth would be the same for all objects, so it is considered to be a constant at the value derived at 10 meters. The result (or more specifically, the conclusion) would be the same either way. I'm holding out hope that you, Aqueous Id, are clever enough to grasp this proof because it appears the others who are (i.e. rpenner) don't have the intellectual fortitude to publicly admit it.
Let me get this straight. You are saying that if I drop a bowling ball from a height of 4.91 meters above the surface of the earth, that it will take 1 second for the ball to impact the surface of the earth?
I reckon 4.91 meters would take 1 second, roughly; unless the mass was vastly larger than a golf ball. (presuming your post is correct! I did not verify)
The heavier item may have a greater attraction but it also has great resistance to change...Intertia: The lighter item has less attraction, and also less Inertia and less resistance to change/movement/falling. So we have one item with great gravitational attraction and more intertia, fighting that attraction...and the other have less gravitational attraction but also less Inertia fighting the gravitational pull. Guess what? The two effects balance each other out, so that they hit the ground at the same time. Galileo was right!
