What if everyone jumps at once at T.F. Green Airport - Peter Tianyi Zhang

You might wonder what would happen if everyone on the planet gathers at a single spot and jumps. Will the orbit of the Earth change? The sum of mass of all people on the planet is huge, and the momentum created by such a huge mass would also be dramatic. However, such a huge mass is still infinitesimally small compared to the mass of the Earth.

Before starting the calculation, there are two questions. First, can all the people in the world fit in T.F. Green Airport? Well, there are about 7 billion people in the world, so it even if everyone stand side by side, it is impossible to fit that much people in such small area as an airport. It is possible, however, to fit all people on the world in an area close to that of Los Angeles, if all standing right next to each other.

Another question is whether any permanent change would be caused by all people jumping together, even if everyone is at the same spot. If we see the Earth and the single spot containing all people as a system, when the momentum and position of the people changes, those of the Earth indeed change, but as the mass of people return to the original spot (back on the ground) after a few seconds, the location and momentum of Earth returns to the original state ,as well, due to the conservation of momentum. But let's just say that we can fit all people in TF Green and we are trying to find the impact within the few seconds all people are above ground.

If we assume that there are 7 billion people, each weigh an average of 80kg, the total mass of people at T.F. Green Airport is

This is a lot, but the mass of earth is 5.972*10^24kg(NASA), which is has a much higher order of magnitude. Let’s assume that each person can jump at an initial velocity of 3m/s vertically above ground, and put this into the conservation of momentum equation. I use E to denote values for Earth and P for values of the sum of human mass.

Sum of momentum, ∑p, always remains constant. In fact, let’s say that the initial sum of momentum is zero, which means at first, nothing is moving. This should be the easiest way to approach the problem, and as we are not concerning anything outside the Earth-people system, it is reasonable to do so.

When people are simply gathered at TF Green, this equation is pretty straightforward: both the velocity of Earth and the velocity of people are zero, so the total momentum is zero.

I find it easier to consider the velocities of Earth and people as functions of time as opposed to value at a specific point of time as we need to look at the duration of time during which all the people jumps. And I plugged in the values for the two masses.

Now suddenly, all the people starts jumping at exactly the same time at an initial velocity of 3m/s

If we organize this function a bit, we can get a function of velocity of Earth vesus time.

So we can graph it now. But we first need to know in what domain is this function valid. I do so by finding the zeros of the kinematic equation of the mass of people

We now know that at 0.6122s, the people return to the ground, and the velocity of people immediately turns to zero. Thus the only part of the function of V_E we are concerned with is between t=0 and t=0.6122.

Intuitively, the velocity of the Earth is negative when the velocity of people are positive, and vice versa, since they are in the same system with a momentum sum of zero. So how much the impact? The  greatest change caused by the jump to Earth’s velocity happens at time 0 and time 0.6122s. At these two point of time, the Earth’s velocity is -2.813*10^13m/s and 2.813*10^13m/s respectively. And how much is that? Well, these velocities are of magnitude 0.2813pm/s, which are extremely small. It seems that 7 billion people’s simultaneous motion has practically no effect on the Earth at all! It is harder to compare this little speed to other things, so let us see how much this motion moves Earth. To do that, we find the integral of the velocity of Earth versus time function

Doing so for the whole 0.6122s of time does not give us the maximum displacement of Earth, as when people return to the ground, the Earth returns to its original position.

A visual representation is the area under the curve of the velocity of Earth vs time function:

Apparently the area of the red negative portion equals the blue positive area so they offset each other. To find the maximum displacement, we must find the highest point of people’s jump, which is midway during the projectile motion. Since the people’s projectile motion starts at t=0, the maximum height of people’s jump is at t=0.6122s/2=0.3061s. Thus the maximum displacement of the Earth is

A visual representation:

So the magnitude of displacement is 4.30581*10^-14m, or 0.0430851pm. How much is that? It is known that the Van Der Waals radius of an hydrogen atom is 120pm (Predictable), which is way larger than the displacement of the Earth calculated above. In fact, the radius on an hydrogen atom is 2785.45 times larger than the Earth’s displacement if this experiment were to be carried out! It can be easily concluded that all people jumping at TF Green Airport at the same time would have practically not effect on the Earth's motion.

Sources:

https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

http://periodictable.com/Properties/A/VanDerWaalsRadius.v.html

http://mentalfloss.com/article/54836/what-would-happen-if-everyone-jumped-once