Might as well jump. Jump. Go ahead, jump. –

Van Halen

Suppose everyone in the world got together and jumped. Would the Earth move? Yes. Would it be noticeable? Time for a calculation. Note: I am almost certain that I have done this before, but I can’t find where.

### Starting assumptions.

- 7 billion people.
- Average weight: 50 kg (you know, kids and stuff)
- Average vertical jump (center of mass): 0.3 meters – and I think that is generous.
- Mass of the Earth: 6 x 10
^{24}kg - Gravitational field near the surface of the Earth is constant with a magnitude of 9.8 N/kg
- Ignore the interaction with the Sun and Moon

### Basic physics

Suppose I take the Earth and the people as my system. In this case, there are essentially no external forces on the system (see assumptions above). There will be two conserved quantities – momentum and energy. Here, the term conserved means that that quantity does not change. I can write:

What does the “1″ and “2″ mean? These could be any two times. For this situation, let me say that time 1 is right after the people jump (and still moving up) and time 2 is when the people are at their highest point.

Energy is also conserved. If I take the people plus the Earth as the system, then I can have both kinetic energy (K) and gravitational potential energy (U_{g}). Using the 1 to represent the people just jumping and 2 to represent them at their highest point, then:

About gravitational potential. First, it is the potential energy of the system, not of each object. Second, in this approximate linear form (mgh), the change is what really matters. This means that I can set the potential at point 1 as 0 Joules. Also, the mass of the Earth does matter in this potential – that is where the 9.8 N/kg comes from.

### The calculation

A couple of important things to start with. At position (and time) number 1, the Earth and the people are moving but there is zero gravitational potential energy. At position 2, the Earth and the people are 0.3 meters apart and not moving (at the highest point). Finally, momentum is a vector – but this is a one-dimensional problem. I am going to let the y-direction be in the direction the people jump.

This gives a momentum conservation equation of:

Now, I can use the energy equation to get an expression for the initial velocity of the people:

Just a quick check with reality. If you want to jump a height h, you would need a speed of:

This is what you get if you assume the velocity of the Earth is super small from above. Ok, I am going to put these two equations (momentum and energy) together. This looks bad, but it really isn’t too bad. The problem is the velocity of the people from the work-energy method still has the velocity of the Earth. Avert your eyes if you are algebra-allergic.

Not finished quite yet – I need to now solve for the velocity of the Earth.

See, that wasn’t too bad. You can open your eyes now. Now for the numbers. If I use the values form above, I get a recoil speed of the Earth as:

Maybe you don’t like my starting values. But you know what? It doesn’t really matter – the mass of the Earth is so huge that it is going to be pretty darn difficult to get a detectable speed. Also, there is the whole issue of getting everyone at the same place at the same time and getting them to jump at the same time.

I seem to recall the last time I did this calculation (that I can’t find) that I also estimated how many people you could get in one spot of the Earth.