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If everyone on earth except one person were to gather up in one location, say the North Pole, and jump at the exact same time and stay airborne for about 1.5 sec. And that one person who did not jump would stand on a weigth scale, would he/she see any difference to their weight assuming the scale can messure weight to a $10^6th$($10^6 $ because I figured that a number with a higher accuracy would change due other changes in the earth or the solar system, such as the gravitational pull of other celestial masses, no?) of a kilogram?

My assumption here is that the force acting on an object from the earth differs if a big mass would suddenly rise from the surface of the remaining mass(the earth), does that make sense? $$F = \frac{Gm_{earth}m_{standing\space person}}{r^2}$$ $F$ would become smaller when $m_{earth}$ becomes decreases due to the jump.


marked as duplicate by Rob Jeffries, ACuriousMind, Brandon Enright, Kyle Oman, JamalS Nov 5 '14 at 7:57

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    $\begingroup$ One answer to this particular question is here what-if.xkcd.com/8 $\endgroup$ – user43617 Nov 4 '14 at 20:24
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    $\begingroup$ A diet would seem a much more easy way to weigh less $\endgroup$ – giulio bullsaver Nov 4 '14 at 20:25
  • $\begingroup$ Yeah, thinking maybe what-if.xkcd should be required reading prior to allowing people to post :-) $\endgroup$ – Carl Witthoft Nov 4 '14 at 20:27
  • $\begingroup$ @Jun-GooKwak, Well, that was disappointing.. $\endgroup$ – Reds Nov 4 '14 at 20:35
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    $\begingroup$ Very closely related: physics.stackexchange.com/q/141973 $\endgroup$ – Kyle Kanos Nov 4 '14 at 20:41

First of all, you have to understand, that even though all these people jumped up from the earth's surface, they would still interact with the standing person (gravitationally).

So, when you say, that earth's mass decreases, you're actually wrong.


It is estimated that the whole population has a mass $316$ million tons: $3.16\times 10^{11}$kg. Earth's mass is estimated to be (I want to take this approximation) $6\times10^{24}$kg. So, taking into account the answer from Chanto and the "insignificance" of the world population mass vs Earth's mass, you can say that the difference would be $0$.


When people jump up, they don't disappear from the earth - they just move by a small amount. Any gravitational attraction there had been before will still be there.

A quick bit of estimating:

7 billion people, each weighing 50 kg (some heavier people in the USA, but lots of children and malnourished adults elsewhere in the world...) - total mass $3.5\cdot10^{11} kg$

Let's say that the lone person stands on the North pole - and notwithstanding all the problems described in the xkcd what-if, they don't all trample each other - but they became a tight circle of friends, 1 meter radius around the person weighing himself.

While they are surrounding the person, there is no net gravitational pull. But when they jump up 1.5 m, there is a "ring of mass" pulling. The distance to every part of the mass is $\sqrt{1.5^2+1^2}$ meter, but some of the forces cancel out. Diagram:

enter image description here

Force scales with $\frac{1}{h^2+r^2}$, vertical component adds another $\frac{h}{\sqrt{h^2+r^2}}$. This makes the net vertical force

$$F_v = G \frac{M_{people} M_{loner} h}{\left(h^2 + r^2\right)^\frac{3}{2}}$$

Putting in the numbers, we get

$$F \approx 400 N$$.

In other words - when you stand that close to an immense amount of mass, the effect it has on you, gravitationally speaking, is substantial. One might even say massive.

But then - if you actually had people standing that close together, the density would put you well on the way to becoming a neutron star (not quite there, but getting there...). So let's put them in a "workable" circle - everyone gets 30x30 cm. Now the size of the disk containing all people has a radius of 14 km, and the disk has a mass density of $50 kg / 0.09 m^2$ = 500 kg/m^2 (we keep rounding... remember we are estimating here)

The problem then becomes computing the gravitational pull of a disk of uniform mass per unit area. It's just an integral. Writing radial distance $r$ and height $h$ as before, we get

$$F = \int_0^R G m 2\pi \sigma r dr \frac{h}{\left(h^2 + r^2\right)^\frac32}$$

Pulling out all the constants, you get (for positive $h$) something of the form

$$\int_0^R \frac{h r\ dr}{\left(r^2+h^2\right)^\frac32}=1 - \frac{h}{\sqrt{h^2 + R^2}}$$

For sufficiently large values of $R$, it really doesn't matter how many people there are... or how high they jump! All that matters is how tightly you pack them on the North pole. They will keep warm.

Putting in the numbers, we get a force of

$$F = 6.67\cdot 10^{-11} \cdot 70 \cdot 2 \pi \cdot 500\approx 15 \mu N$$

This is something that could be measured, but not on a bathroom scale.

Standing around on the North Pole you will be getting cold, which causes you to lose weight quickly. Also, because you are closer to the center of the earth AND not experiencing the benefit of the earth's rotation on your apparent weight, you will actually be quite a bit heavier... about 0.5% more than at the equator, of 3.5 N for the same 70 kg person. Now THAT is something you can measure. Even on good bathroom scales.

Incidentally, if you were standing on a different part of the earth - further from the people jumping - the effect would be much smaller... That's the inverse square law for you (or against you, depending on your point of view).


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