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What if everybody in the world jumped at the same time at the same location?  It is practically impossible for all people to be in the same location at the same time (mainly because there are 7.6 billion of us).  Although one may think that the mass of all the people in the world combined may be a lot (which it is), it is nowhere close to the mass of the world.  Due to the major difference in masses, everybody jumping at the same time would have no affect on the world at all.

Values that I used:
Amount of people of the world: 7.6x10^9
Average mass of a person: 70kg
Average vertical jump of a person: .5m
Mass of the Earth: 5.972 x 10^24 kg

An elastic equation will be used because momentum is conserved, which means kinetic energy is also conserved.  The equation that I used is the conservation of momentum equation.

m1v1+m2v2=m1v1f+m2v2f

To solve for the velocity of the earth I used a conservation of energy equation.  H in this equation is equal to zero because the Earth did not move vertically.

1/2m1v1^2 + 1/2m2v2^2 = mgh
1/2m1v1^2 + 1/2m2v2^2 = 0
1/2m1v1^2 = -1/2m2v2^2
m1v1 = -m2v2
v1 = m2v2/m1

I then used this same energy equation to calculate the velocity of the people.  H in this equatoin is not equal to zero, and the mass in the potential energy is m2.

1/2m1v1^2 + 1/2m2v2^2 = m2gh
m1v1^2 + m2v2^2 = 2m2gh
m2v2^2 = 2m2gh-m1v1^2
v2^2 = (2m2gh-m1v1^2)/m2
v2^2 = 2gh - m1v1^2/m2
v2 = sq. rt. 2gh - m1v1^2/m2

Putting both equations together allowed me to find the velocity of the earth.

v1^2 = (2ghm2^2/m1^2) - (m2v1^2/m1)

Solving for v1 will give me the velocity of the earth.

v1^2 + (m2v1^2/m1) = (2ghm2^2/m1^2)
v1^2 (1 +m2/m1) = (2ghm2^2/m1^2)
v1^2 = 2ghm2^2 / m1^2 (1 +m2/m1)
v1^2 = 2ghm2^2/ (m1^2+m2m1)
v1 = sq. rt. 2ghm2^2/ (m1^2+m2m1)

The mass of the total population of the world is equal to:
(7.6 x 10^9)*70 = 5.32 x 10^11 kg.

v2 = sq. rt. 2*9.8* (5.32 x 10^11)^2 / (5.972 x 10^24)^2 + (5.972 x 10^24)*(5.32 x 10^11)
v2 = (2.17589 x 10^24)/ (3.56648 x 10^49)
v2 = 7.94 x 10^-27 m/s

This means that the velocity of the Earth will only be affected by 7.94 x 10^-27 m/s after the jump.  The change is so small because the mass of the entire human population is nowhere close to the mass of the earth.  






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