What if exactly half the Earth's population jumped at one instant? + Secondary Question I read somewhere that when you jump, the sole effect caused by your jump on the earth moves it about $10^{-18}m$ (I don't remember the figure exactly, but I think it was that).  
However - obviously - with so many people running, jumping, etc.  on the whole surface of the earth, the whole effect is canceled out, so it's safe to say there is no net displacement caused by us.
This got me to wonder, what if the whole population of one longitudinal hemisphere of the earth jumped at the exact same time - while the other half remained stationary - would we be able to notice any discernable displacement in the short time before we all land back to the ground?
Let's consider the earth to have a perfect distribution of people on both hemispheres, and let's consider the people of the Eastern Hemisphere to jump. Now the people jumping along the Prime Meridian and the International Date Line (these two lines together form the circle that divides the earth into the West and Eastern Hemispheres) would produce no effect, since they would simply cancel each other out. However, the closer you go to $0°N, 90°E$ (the centre of the East Hemisphere), the component of force which adds up will keep increasing. So there will be a net force in the direction of the Western Hemisphere. I was wondering if it would be noticeable. 


Another question which just struck me: 
Out of all the satellites/rockets/space-shuttles we've sent out, have they caused any displacement on the earth? They are quite massive in comparison to us(not the earth, of course), and have large speeds. Many of them are continuously moving further and further away from the earth - so there's no question of them falling back to earth and thus shifting it to its original position again. Will the momentum be enough to displace the earth?
 A: Some rough figures: earth's mass is about $6 \cdot 10^{24}$ kg. The mass of the total world population is roughly 7 billion times 80 kg or about $6 \cdot 10^{11}$ kg. So earth is 13 orders of magnitude (10 trillion times) heavier than the world's human population.
Suppose that the total mass of people gets together at one spot and everyone jumps up at the same time with a speed of about 1 m/s. The center of mass of earth plus all jumpers continues its trajectory, but momentum (mass times velocity) conservation dictates a recoil speed of $10^{-13}$ m/s for earth relative to the center of mass. That speed lasts for about a second and gets reversed due to gravitational attraction. When everyone lands on his/her feet earth is back at its normal trajectory.
The displacements (compared to the center of mass trajectory) follow the same mass ratio: if all the jumpers reach a respectable height of 1 m, earth gets displaced by $10^{-13}$ m (1/500th of the diameter of a hydrogen atom).
Distributing everyone over a hemisphere would reduce the effect by some factor. Hardly relevant, as there is no significant effect anyway.
A: 
Discernable displacement?

No, the earth far outweighs the human population.  If the masses were closer, absolutely, as action/reaction.
More interestingly is the effect of Gravity on the separate masses.  If both masses lack sufficient "escape velocity", both return to their original positions.  If the original "push" has escape velocity, we continue on into space and the earth goes slightly faster in the other direction (recoil).
But for chemical rockets, the earth is actually pulled towards the rocket and will move slightly towards it.  The chemical action replaces "push" towards the earth as the driver of propulsion.
Gravitational "towing" has been proposed for altering the path of earth crossing asteroids.
