Earth Will Drop Downwards?


Kao-Kan's avatar
If everybody on Earth went to the North Pole and jumped up and down at the same time, would the weight of it all cause the earth to drop downwards in space?
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FixMeKnow's avatar
Your avatar looks really familiar. Why is this? :evileye:
ScottaHemi's avatar
It won't work like that, you need something big moving at very fast speed to shift the orbit.

though in theory if you concentration everyone at the equator the earth may slow it's rotation a bit due to centrifugal force.
Urus-28's avatar
Well I don't know for earth, but for the ice of the north pole it would break and kill all humanity in the cold waters. :fuzzydemon:
hopeira9's avatar

The earth shall do whatever you want it to... if you're PinkiePie
KoutaTheCreator's avatar
:icongo--onplz: Yes, of course it would.
Vanhir's avatar
No, because the polar bears wouldn't allow it.
Endeavor-To-Freefall's avatar
Also the idea that we revolve sideways around the sun is false, a more correct way to think about it is that we're constantly falling downwards. After all, if the sun is pulling us inwards, what would be holding us up?
Gemdrop's avatar
But nothing falls in space..?
Endeavor-To-Freefall's avatar
It's just a matter of perspective, when we drop objects they fall to the floor because gravity pulls them towards it. On a larger scale, the Earth is falling when gravity pulls it towards the Sun. It's just moving at enough velocity that it never actually touches it.
Juliabohemian's avatar
This is why I don’t do drugs.
Kao-Kan's avatar
Likewise. Just pops of the lolly-ish variety.
Endeavor-To-Freefall's avatar
Gravitational Energy E= -mu/r + mcV , where -mu/r is the Potential Energy and
mcV is the Vector Energy.

Continuity of E gives :

0=(d/dr + Del)m(-u/r + cV) = m(v^2/r - cv/r cos(x)) + m(cdV/dr -Del u/r).

The real part is zero and is tge balance of centripetal force mv^2/r and centrifugal force mcDel.V = mcv/r cos(x):

mv^2/r - mcv/r cos(x) =0 so mv^2/r = mcv/r cos(x) gives v/c= cos(x) =z the redshift and equilibrium.

The vector part is also zero and gives the orbit:
Fv= (dcV/dr + u/r^3 R + cDelxV )

Enough weight could alter our orbit but it would have to come from outside it, and would simply change our course a tiny amount, what we gain moving toward the sun would then propel us slightly further away on the other side, distance traveled would ultimately remain the same. It wouldn't work for people jumping though, if you've ever heard of the chickens in the truck it's the same principle. Our weight is already accounted for when we're within Earths gravitational field, if we were to jump, the mass of the Earth stays the same, the force of our jump and our landing ultimately cancel each other out. It would cause a pretty big earthquake though. It could possibly alter our axis.
Kao-Kan's avatar
Jumping energy (J+umping energy = Jumping energy (+) north pole = E+arth falling downwards through s+pace. Accept it.
Endeavor-To-Freefall's avatar
shininginthedarkness's avatar
1. There is no 'down' in space

2. Compared to the weight of the planet, people are nothing. This would be like a flea jumping up and down on my head and knocking me out of my chair.
Kao-Kan's avatar
Of course there is a 'down' in space. It's underneath the planet.
shininginthedarkness's avatar
Everywhere is underneath the planet!
Kao-Kan's avatar
We will have to agree to disagree on this one. :(