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`

Step-by-step explanation:

The equation to calculate the center of mass
`C_(M)` of a particle system is:

`C_(M)=(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)=(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)=(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.