A 90 kg person stands on the ground before jumping vertically upward. If momentum is truly conserved in all closed isolated systems, then the Earth-person system should conserve momentum. If the person jumps with a speed of 3.5 m/s, what is resulting speed of the Earth in the opposite direction? (Earth's is approximately 5.96*10^24 kg)
A person with mass 90 kg standing on the Earth jumps vertically upwards. The speed of the person is 3.5 m/s. The Earth and the person standing on it can be considered as a closed system. The law of conservation of momentum is always true.
The momentum of the person when he jumps upwards is 90*3.5 = 315 kg*m/s. This should be equal to the momentum of the Earth in the opposite direction. Let the resulting speed of the Earth be v. As the mass of the Earth is equal to 5.96*10^24 kg, v*5.96*10^24 = 315
=> v = 315/(5.96*10^24)
=> v `~~` 5.285*10^-23 m/s
The resulting speed of the Earth is 5.285*10^-23 m/s