Suppose the entire population of the world gathers in one spot and, at the sounding of a prearranged signal, everyone jumps up. While all the people are in the air, does Earth gain momentum in the opposite direction?

As you know, the Law of conservation of momentum states that the total momentum does not change for a closed system.

That is to say that if you're dealing with a system that's isolated from the exterior, meaning that you get no external forces acting upon it, then a collission between two objects will always result in the conservation of the total momentum of the system.

The total momentum is simply the sum of the momentum before the collision and the momentum after the collision.

Now, if you take Earth to be a closed system, then the momentum of the Earth + people system before the people jump must be equal to the momentum of the Earth + people system while all the people are in the air.

From the perspective of the Earth, it is important to understand that once the people land back on the surface, the momentum of the Earth will be the same as it was before they jumped.

So, let's assume that the initial momentum of the Earth + people system was zero.

If all the people jump at the same time, then the combined mass of the jumpers, #m#, will have a velocity #v_"people"#, and a momentum of #p_"people"#.

This means that in order for the total momentum of the system to be conserved, the Earth, let's say of mass #M#, will need to have a velocity #v_"Earth"#, and a momentum oriented in the opposite direction to that of the people.

#overbrace(0)^(color(blue)("momentum before the jump")) = overbrace(p_"people" + p_"Earth")^(color(green)("momentum after the jump"))#

This is equivalent to

#0 = m * v_"people" - M * v_"Earth"#

The minus sign is there to show that the velocity of the Earth is oriented in the opposite direction to that of the people.

However, the difference between the mass of the Earth and that of the people will make this change in momentum very, very, very small.

A quick calculation to illustrate that. Let's take the mass of the Earth to be #6.0 * 10^(24)"kg"#. Assuming an average weight of #"60 kg"# per person and a total of 7 billion people, you'd get

#m * v_"people" = M * v_"Earth"#

#v_"Earth" = v_"people" * m/M#

#v_"Earth" = v_"people" * (60 * 7 * 10^9color(red)(cancel(color(black)("kg"))))/(6.0 * 10^(24)color(red)(cancel(color(black)("kg"))))#

#v_"Earth" = 7.0 * 10^(-14) * v_"people"#

The velocity of the Earth will be smaller than that of the people by a factor of #7 * 10^(-14)#.