Problem — Jumping out of the window

Problem

A person of $75\;\text{kg}$ jumps from $h = 5\;\text{m}$. Treating the person-plus-Earth system as an “explosion” (gravitational potential energy transforms into kinetic energy), estimate the recoil velocity of the Earth and comment on why we do not notice it. Mass of the Earth $M_\oplus \approx 6\cdot 10^{24}\;\text{kg}$.

Solution

Velocity of the person at the end of the fall (conservation of energy, neglecting the tiny kinetic contribution of the Earth): $\tfrac{1}{2}m v_1^2 = m g h \quad\Rightarrow\quad v_1 = \sqrt{2gh} = \sqrt{2\cdot 9{,}81\cdot 5} \approx 9{,}9\;\text{m/s}$

Conservation of $\vv{p}$ (person + Earth system initially at rest, like a two-fragment explosion): $0 = m v_1 + M_\oplus v_\text{Terra} \quad\Rightarrow\quad v_\text{Terra} = -\frac{m\,v_1}{M_\oplus}$ $v_\text{Terra} = -\frac{75\cdot 9{,}9}{6\cdot 10^{24}} \approx -1{,}2\cdot 10^{-22}\;\text{m/s}$ $\ev{v_1 \approx 9{,}9\;\mathrm{m/s},\qquad v_\text{Terra} \approx 1{,}2\cdot 10^{-22}\;\mathrm{m/s}}$ Comment. The Earth’s momentum is equal and opposite to the person’s, but its mass is $\sim 10^{23}$ times larger: the recoil velocity is imperceptible, $\sim 10^{-22}\;\text{m/s}$, many times smaller than the diameter of an atomic nucleus per second. This is why the Earth seems like a stationary reference frame to us.

Links

Topics: Quantità di moto e urti Concepts: Conservazione della quantità di moto · Esplosione · Energia potenziale gravitazionale Skills: Conservazione della quantità di moto