this post was submitted on 18 Sep 2025
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https://www.usgs.gov/water-science-school/science/how-much-does-a-cloud-weigh
Planes, helicopters- lots heavy stuff not falling faster than lighter ones
Depends on whether or not you count in air resistance. I was just making a shitpost
Interesting way to admit you were wrong
You can find exceptions, but on average, heavier objects will fall very slightly faster than light ones, because they excert their own gravity field onto Earth and therefore pull it towards themselves.
This requires a somewhat unintuitive definition of "falling", in that both the object and Earth itself moves, but given that any object with mass excerts a gravitational field, there is not actually any other definition.
No.
https://en.m.wikipedia.org/wiki/Equivalence_principle
Wut? This does not turn off gravitational pull for objects other than Earth.
Or I'm misunderstanding what you're trying to say, but yeah, no clue.
You didn't read it, it is literally telling you you are wrong.
Ah, I'm not saying there's a different force being applied to feather vs. hammer. The meme above doesn't mean that they "fall faster" in the sense that the hammer falls at a higher velocity. It's rather colloquial usage of "faster" to mean "finishes sooner". Because what does happen, is that the hammer collides sooner with Earth, since the hammer pulls the Earth towards itself ever-so-slightly stronger than the feather does.
I guess, for this to work, you cannot drop hammer and feather at the same time in the same place, since they would both pull Earth towards themselves with a combined force. You need to drop them one after another for the stronger pull of the hammer to have an effect.
So, this is also going off of this formula:
But setting
mass_1
as Earth's mass andmass_2
as either the feather's or hammer's mass. A highermass_2
ultimately leads to a higher force of attractionF
.So in that equation, let's say mass 1 is earth. G and distance will be equal in both instances of dropping.
Rewrite equation:
Distance^2/ G*mass 1 = mass 2 /force
And
Distance^2/ G*mass 1 = mass 3 /force
Therefore,
Mass 2 /force = mass 3 /force
F = m*a
Mass 2 / mass 2*a = mass 3 / mass 3 * a
This cancels out to show that a = a, their acceleration is the same.