- Jan 15, 2001
- 15,069
- 94
- 91
fall from an equal point above the earths surface. if ball 1 is 100000x heavier than ball 2, does ball 1 fall faster?
edit:
thanks to dullard, here is a link to NASA's thought on the matter...
Text
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proof using math from about 40 posts down
F=ma, a=32ft/sec, m= whatever the mass of the object is. The acceleration is always the same IF you ignore air resistance.
If you don't ignore air resistance, then the shape of the object has to be taken into account. The shapes were given as 2 spheres of equal volume. So the shape is out as far as affecting the rate at which they will fall.
All that's left is the mass of each object being different.
Since the force on the objects can be shown as F=ma without air, with air resistance it would be F=ma-r (r stands for resistance (air resistance but I can't write ar or you'll think it's acceleration times r, which it's not)
F=ma-r, a =g, the constant for gravity
so therefore F=mg-r, and mg-r=ma
solve for a, you get a= g-r/m (divided both sides by m)
this gives us what the acceleration will be with wind resistance.
as you can see, if the mass is large, r/m becomes small and wind resistance won't affect it much. but if m is small, the 2nd term r/m starts to become significant. This will "subtract" more from the acceleration. get it?
==============
the last part makes perfect sense if you read it carefully...
everyone down below is right...this will be debated to no end like the other thread where people tried to say .99999 doesnt = 1 (which it does...if you dont think so, refer to basic calculus and stfu)
edit:
thanks to dullard, here is a link to NASA's thought on the matter...
Text
==============
proof using math from about 40 posts down
F=ma, a=32ft/sec, m= whatever the mass of the object is. The acceleration is always the same IF you ignore air resistance.
If you don't ignore air resistance, then the shape of the object has to be taken into account. The shapes were given as 2 spheres of equal volume. So the shape is out as far as affecting the rate at which they will fall.
All that's left is the mass of each object being different.
Since the force on the objects can be shown as F=ma without air, with air resistance it would be F=ma-r (r stands for resistance (air resistance but I can't write ar or you'll think it's acceleration times r, which it's not)
F=ma-r, a =g, the constant for gravity
so therefore F=mg-r, and mg-r=ma
solve for a, you get a= g-r/m (divided both sides by m)
this gives us what the acceleration will be with wind resistance.
as you can see, if the mass is large, r/m becomes small and wind resistance won't affect it much. but if m is small, the 2nd term r/m starts to become significant. This will "subtract" more from the acceleration. get it?
==============
the last part makes perfect sense if you read it carefully...
everyone down below is right...this will be debated to no end like the other thread where people tried to say .99999 doesnt = 1 (which it does...if you dont think so, refer to basic calculus and stfu)