Question

(a) How far is the center of mass of the Earth-Moon system from thecenter of the Earth? (Appendix C gives the masses of the Earth andthe Moon and the distance between the two.)

(a) How far is the center of mass of the Earth - Moon system from thecenter of the Earth? (Appendix C gives the masses of the Earth andthe Moon and the distance between the two.) (b) Express the answer to (a) as a fraction of the Earth'sradius. (Rcm / Rearth) =
m

(b) Express the answer to (a) as a fraction of the Earth'sradius.
(R_cm / R_earth) =

Answer 1

Concepts and reason

The concept used in this question is center of mass of a system. The standard values for mass of Earth, mass of Moon, radius of Earth and distance of Moon from the Earth are used here.

First, find the position of center of mass of the system using standard values and then find its ratio with radius of the Earth.

Fundamentals

Center of mass:

The center of mass of a system is a point where all the mass of the system is concentrated. It is also known as the center of gravity because the force due to gravity acts at this point.

It is calculated about an axis. If there are two objects in a system then, the formula for position of center of mass of the system is:

${x_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$

Here,

$\begin{array}{l}\\{m_1}{\rm{ and }}{m_2}{\rm{ are masses of objects}}{\rm{.}}\\\\{x_1}{\rm{ and }}{x_2}{\rm{ are their positions from the axis}}{\rm{.}}\\\end{array}$

Standard values:

The mass of Earth is $5.972 \times {10^{24}}\;{\rm{kg}}$

The mass of Moon is $7.34767 \times {10^{22}}\;{\rm{kg}}$

The radius of Earth is $6371\;{\rm{km}}$

The distance of Moon from the Earth is $384400\;{\rm{km}}$

(a)

The position of center of mass of the system from the center of Earth is:

${x_{cm}} = \frac{{{M_{earth}}{x_1} + {M_{moon}}{x_2}}}{{{M_{earth}} + {M_{moon}}}}$

Here,

$\begin{array}{l}\\{x_1}{\rm{ is the distance of Earth from the Earth i}}{\rm{.e, 0}}\\\\{x_2}{\rm{ is the distance of Moon from the Earth i}}{\rm{.e, }}384400\;{\rm{km}}\\\end{array}$

$\begin{array}{c}\\{x_{cm}} = \frac{{{M_{earth}}{x_1} + {M_{moon}}{x_2}}}{{{M_{earth}} + {M_{moon}}}}\\\\ = \frac{{\left( {5.972 \times {{10}^{24}}\;{\rm{kg}}} \right)\left( 0 \right) + \left( {7.34767 \times {{10}^{22}}\;{\rm{kg}}} \right)\left( {384400\;{\rm{km}}} \right)}}{{\left( {5.972 \times {{10}^{24}}\;{\rm{kg}}} \right) + \left( {7.34767 \times {{10}^{22}}\;{\rm{kg}}} \right)}}\\\\ = \frac{{2.824 \times {{10}^{28}}\;{\rm{km}} \cdot {\rm{kg}}}}{{6.045 \times {{10}^{24}}\;{\rm{kg}}}}\\\\ = 4671.6\;{\rm{km}}\\\\ \approx {\bf{4672}}\;{\bf{km}}\\\end{array}$

(b)

The position of center of mass in terms of radius of Earth is:

$\begin{array}{c}\\\frac{{{x_{cm}}}}{{{R_{earth}}}} = \frac{{4672\;{\rm{km}}}}{{6371\;{\rm{km}}}}\\\\ = 0.733\\\\{x_{cm}} = {\bf{0}}{\bf{.733}}\;{R_{earth}}\\\end{array}$

Ans: Part a

The position of center of mass of the system is 4672 km.