Question

Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one place and climb to the top of 2 m high ladders. How muchhas the center of mass of the Earth (mE = 5.90×1024 kg) displaced as a result?

Answer 1

Answer: $5.9(10)^{-8} m$

Explanation:

The equation to calculate the center of mass $C_{M}$ of a particle system is:

$C_{M}=\frac{m_{1}r_{1}+m_{1}r_{1}+...+m_{n}r_{n}}{m_{1}+m_{2}+...+m_{n}}$

In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:

$C_{M}=\frac{m_{E}(0 m) + m_{p} r_{E-p}}{m_{E}+m_{p}}$

Where:

$m_{E}=5.9(10)^{24} kg$ is the mass of the Earth

$m_{p}=55(10)^{9} kg$ is the mass of 1 billion people

$r_{E}=6371000 m$ is the radius of the Earth

$r_{E-p}=6371000 m- 2m=6370998 m$ is the distance between the center of the Earth and the position of the people (2 m above the Earth's surface)

$C_{M}=\frac{m_{p}55(10)^{9} kg (6370998 m)}{5.9(10)^{24} kg+55(10)^{9} kg}$

$C_{M}=5.9(10)^{-8} m$ This is the displacement of Earth's center of mass from the original center.