Sorry a typo F = 5.88e25 not 5.88e17
Gunnar Valur Gunnarsson
There’s some poor physics above, and some misunderstandings. The acceleration of the hammer and the acceleration of the feather will be the same. The acceleration of either depends does not depend on their masses. What the original question ( http://www.straightdope.com/mailbag/mgravity.html ) asks is whether or not the time is takes for each object to hit the earth is the same. It is not exactly identical, but is close enough for anyone but a stubborn philosopher (perhaps I’m one of them).
If you really want to know how long it takes two freely moving objects to collide, here’s the answer. Not for the math impaired. Setup: two objects at rest with respect to each other and only experiencing their mutual gravitational attraction. Without loss of generality we can assume they are spheres and consider motion in only one dimension. We’ll give the masses m1 and m2, radii r1 and r2, positions x1 and x2. Initial conditions are initial distance is d=x2(0)-x1(0) and initial velocities v2(0)=v1(0)=0.
The Langrangian for this system is L=(1/2)(m1v1^2+m2v2^2)+Gm1m2/(x2-x1). From here we can derive the equation of motion (and I have done so correctly): (d2/dt2)x=-GM/x^2, where x=x2-x1 (the distance between the objects) and M=m1+m2 (the total mass). If x does not change much and one mass is much larger than the other, this will reduce to the constant acceleration approximation we’ve been using until now. The solution to this equation with the given initial conditions is t=sqrt[d^3/(8GM)](2sqrt[u*(1-u)]+arctan[2*sqrt[u*(1-u)/(u-(1-u))]). The total time until collision is t, and u=(r1+r2)/d. Be careful to pick the correct quadrant when you calculate the arctangent.
Now anyone(?) can calculate how long it takes something to fall. For the same situation I gave in my earlier post, the difference in time between a 2kg hammer and 1g feather is 7.54e-26 seconds. Of course this is ignoring that the earth and the objects do not start at rest with respect to each (e.g., the earth’s rotation), but the corrections will not greatly change the answer.
Sorry for the math overkill, but there are too many misunderstandings arising from the approximations many of the posters are making implicitly.
Yeah, but what if you drop the feather and fire the hammer out of a gun at the same time?
That would be orbital mechanics–you’re no longer guaranteed the hammer will hit the earth.
Thanks Pleonast
If someone didn’t understand what I said in my original post, ( I see that you did ) they can be quite assured now that what I was trying to say is true and you have proven it. You have also shown that the difference is very small so we can easily ignore it but we shouldn’t since this is theory not practis.
Gunnar Valur Gunnarsson
I think that much of the difficulty we are having here arises from a failure to specify the coordinate system in which we are measuring acceleration. Keep in mind that Newtonian mechanics only applies in an inertial reference frame. Given that the question allows for acceleration of the Earth, the conventional choice of an Earth-fixed coordinate system is inappropriate and will yield bizarre results.
I would suggest that the best coordinate system to choose is one with origin fixed at the center of mass of the combined system. In this coordinate system, the acceleration on each particle due to the others is independent of its own mass, depending only on the masses of and distances to the other objects. This point has been demonstrated several times. The acceleration computed is not the acceleration with which two particles move toward each other; it is the acceleration towards the system CM.
Now, there is another point of confusion. Are we talking about simultaneous arrival at the CM, or simultaneous arrival at the surface of the earth? If we’re talking about simultaneous arrival at the CM, then (in a system with zero initial kinetic energy and angular momentum…) all particles arrive at the CM simultaneously. If you don’t want to do the detailed calculation, think of it this way: if one particle got to the CM first, it wouldn’t be the CM anymore. If you do want the full calculation, keep in mind that, although all particles accelerate at different rates, they have different distances to traverse.
But the original question is, which hits the *ground[/] first? Leaving out issues like the fact that the Earth isn’t spherical and doesn’t have uniform density and making the hammer and feather point particles and so forth, in this case, I think the hammer always has the edge, but I’m hard-pressed to prove it. We’re talking three-body problem, here, folks, and it’s famous for not having a nice general solution like the two-body problem. If you admit of the rotation of the earth, then it’s really complicated since the three particles are in orbit around each other after release of the hammer and feather. (This is why they launch rockets east from Cape Canaveral.)
To me, though, the really interesting question is this: why is it that everyone is perfectly willing to cancel the mass that appears in Newton’s Second Law (F=ma) with the mass in Newton’s Universal Law of Gravitation (F=GmM/r^2)? Why do you think that inertial mass and gravitational charge are mathematically identical?
–Y
``Is inertial mass identical to gravitational mass?’’, that is the fundamental question everyone is asking (even if they don’t realize it). Newton’s Laws of Motion and Gravity and both of Einstein’s Theories of Relativity assume they are mathematical identical. Experimentally, the answer is that our best measurements cannot find any difference to within error bonuds (about one part in 10^10). These experiments carefully compare accelerations of differenct objects under gravity. Simply put, there’s no reason to think inertial and gravitational masses are different.
BTW, there is a typo in the time formula of my previous post. Both sqrt’s have the same argument: [u*(1-u)]. I left out the closing bracket of the second root. And, I should mention that my Lagrangian dynamics is wholly classical: no special or general relavity and no quantum mechanics.
[quote]
Simply put, there’s no reason to think inertial and gravitational masses are different.
[quote/]
Well, the experimental evidence strongly indicates that they are identical, anyway. The question I was raising, however, is that of *why[i/] they should be identical when they arise from two *different[i/] phenomena–inertia and gravitation. The fact that gravitational charge and mechanical inertia appear to be exactly equal indicates that gravitation and inertia are at least related, or even the same phenomenon. Welcome to General Relativity! And to think we started with feathers and guineas (errr… hammers). Ain’t science great?
Newton did not merely assume that inertial mass and gravitational mass were identical; he observed that both the two-weight experiment and Keplerian orbital mechanics were explained thereby.
General Relativity is partly a result of Einstein asking, “OK, so they are identical, but why?”
John W. Kennedy
“Compact is becoming contract; man only earns and pays.”
– Charles Williams
And Einstein’s conclusion was that the “force” of gravity is as fictitious as centrifugal “force” or the Coriolis “force.” What we perceive (and measure) as the force of Gravity is just inertia–the simplest possible physical law: things don’t change if you don’t do anything to them. Orbiting objects coast along straight lines in curved space. What a concept!
I found this thread from this newer thread, and I was surprised at the ignorance of physics.
AuraSeer posted 09-13-1999 08:41 AM
First of all, 9.8 + 9.8 =19.6 != 9.8 Second of all, in the object frame, the Earth accelerates at 9.8, and the object is stationary. In the Earth frame, the Earth is stationary, and the object is accelerating at 9.8. In no frame are both accelerating towards each other at 9.8. Pick a frame and stick with it!
Are you saying that a hammer with the mass of the Earth would have less acceleration than a normal hammer? Why?
Why is the Earth so special? Why does the acceleration depend on the mass of the Earth and not on the mass of the object? Is the Earth made of some special substance?
HeadlessCow posted 09-13-1999 10:36 PM
It gives both the force on Earth and the hammer. To find the acceleration, you have to divide by the appropiate mass.
Wat? The more massive two objects are, the more they attract. Increasing the mass of the Earth increases the acceleration of the object (but not the acceleration of the Earth, at least not directly). Increasing the mass of the object increases the the acceleration of the Earth.
As for the original question: since the term “hitting the Earth” refers to the time at which the valence shells of the Earth and the object would start to have significant interference, and the difference would be about a dozen orders of magnitute smaller than the valence shell diameter, there would be no detectable difference.
If you really want the exact difference, you would have to solve a three-body problem. The frist order answer, I suppose, would be that the mass of the two objects has absolutely no effect. The second order answer would be that the more massive the objects are, the faster they fall, but since they’re dropped at the same time, they’ll hit at the same time. I suppose as a third order answer we can take the mass of the feather to be negblible. Then Pleonast’s formula would apply to the motion of the hammer and earth, but not the feather. For the feather, set the inititial condition to be velocity=(0,0,0), position=(2 meters, arm’s length, 0). Then use the hammer and Earth’s positions to calculate the force at each intsnat, and integrate. I don’t think can be done in any manner other than numerically, and I don’t have access to a program that can keep track of all the significant digits necessary to get any answer other zero (I suppose I could write my own program, but it doesn’t seem all that important). My guess is that seeing as how the horizontal separation of the hammer and feather is about four order of magnitudes smaller than the radius of Earth, this will lead to an answer four orders of magnitude smaller than Pleonast’s answer (assuming, of course that it’s correct).
Ok, if after 17 years, and before that Apollo 15, and before that that Italian guy and math, here’s a demonstration in the largest vacuum “chamber” on good old Terra Firma: http://interestingengineering.com/video/happens-bowling-ball-feather-dropped-vacuum/
If you’re in a hurry, it starts getting interesting at 2:00 in, and The Proof is experimentally demonstrated at 2:41.
Neat facility.
I thought everyone knew that a ton of feathers is heavier, because you also have to carry the weight of what you did to all those poor birds.
Whoever’s bowling ball that is is gonna be pissed …
<nitpick>Galileo used an incline plane, he didn’t just drop the weights {Cite} </nitpick>
There are two things you can learn from watching that video. The feather hit the ground first, and second, things fall really slowly in a vacuum.
Galileo didn’t have a high-frame-rate camera.
I ain’t in the mood for any complicated equations. But I can assure you that if the earth and a hammer of earth mass approached each other in a gravitational death spiral, the differential densities of the components of the earth and of the hammer would very quickly become relevant. So would the different between being shaped as a near-sphere and as a hammer.
It would be quite likely that denser interior portions of the earth would move through less dense portions in their quest to meet the mega-hammer.
We’d even change our definition of “earth”. In our standard feather/hammer calculation, even though we like to ignore air resistance/surface area-to-mass distribution factors, despite the immense importance of this to the flight capacity of the creatures supplying the feathers, we’re still, astronomically speaking, inside the earth when we are dropping things.
But in an astronomical-scale collision, the atmosphere is much more obviously very much part of the earth, and an important part, and one very much affected by the gravitational attraction of large objects, in comparison to other, denser, components of the earth.
Ultimately, though, the atmosphere is the answer to the original question. If all objects that, astronomically speaking were sufficiently close to earth, approached the earth with equal speed regardless of mass, there wouldn’t be an atmosphere.
The fact is that it is only because the difference in the masses of the two objects “dropped” upon the earth is negligible compared to the mass of the earth that we can pile another fiction about mass upon our fictions about air resistance and density.
But relative mass is extremely important in determining whether a large object moving at a given speed at a given distance will have a collision with the earth or not, and if so, whether that collision sucks away some of the atmosphere, some of the crust, or results in a full liquidation of both objects followed by successor objects that probably deserve different names.
Like “Theia” becoming “Luna”.
Whichever one is measured in metric tons.
Isn’t this basically what LIGO detects? Stand on one black hole, pick up second equally massive black hole, drop hole with enough kick that it spirals in.
So if you had a interferometer the size of (something largish like a universe), drop the hammer from the ISS and you can measure the waves as it falls toward Earth.