![]() Thus the smaller object's net force is further reduced proportional to mass at a given velocity. For the smaller object: 5-2=3 (60% of the gravitational force). For the heavier object, the net downward force is 10-2=8 (80% of the gravitational force). the heavier object has, for example, a downward force of 10, the lighter object 5, and the upward (resistive force due to air) is 2 at a given velocity (who cares about units?). It seems to me that we would then be discussing force subtraction, so relative magnitudes do matter. *perhaps there's a question still remaining (for me, and perhaps others): is acceleration still equal in the presence of air-resistance up to the point of terminal velocity? Or does air resistance, being an opposing force to gravity, reduce acceleration slightly right 'from the get-go'? The sum of gravity and air resistance (which is independent of mass) is not proportional to mass. ![]() Things fall the same way if the net force (sum of all forces) is proportional to mass: Only in a uniform gravitational field would these two hit at the same time. We let go and the dense one hits first because its center of mass is lower and that increases the pull of gravity on it due to it being on average at a lower radius. ![]() If you want more titles like this, then check out Happy Glass or Happy Glass 2. We put two spheres with the same mass but different density on a plate a km up. Physics Drop is one of our handpicked puzzle games that can be played on. Each travels a different distance because the Earth is not sitting still. ![]() It accelerates more or less at the same rate at first, but gains speed quicker because the Earth is accelerating up towards the massive moon but not the balloon moon. Now we let go of something the mass of the moon (and stronger to avoid Roche effects) from the same altitude and it will hit the ground in significantly less time. We let go of it and time how long it takes to hit. Put your brain to work in this addictive puzzle game that millions of players already enjoyed in mobile platforms and now is finally ready for PC. We have a controlled Earth (not spinning, a perfect sphere, no air) all alone in space (so the sun isn't accelerating it for instance) and I have a stationary moon-size rigid balloon at the same altitude as the orbit of the moon. Assume m2 masses a significant percentage of Earth, like the moon say. OK, since we're getting picking about details, if you do the computation to enough significant digits, a dense object will fall faster as will a more massive hinted at this. ![]()
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