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9.6 Center of mass  (Page 4/13)

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Here are two examples that will give you a feel for what the center of mass is.

Center of mass of the earth-moon system

Using data from text appendix, determine how far the center of mass of the Earth-moon system is from the center of Earth. Compare this distance to the radius of Earth, and comment on the result. Ignore the other objects in the solar system.

Strategy

We get the masses and separation distance of the Earth and moon, impose a coordinate system, and use [link] with just N = 2 objects. We use a subscript “e” to refer to Earth, and subscript “m” to refer to the moon.

Solution

Define the origin of the coordinate system as the center of Earth. Then, with just two objects, [link] becomes

R = m e r e + m m r m m e + m m .

From Appendix D ,

m e = 5.97 × 10 24 kg
m m = 7.36 × 10 22 kg
r m = 3.82 × 10 5 m .

We defined the center of Earth as the origin, so r e = 0 m . Inserting these into the equation for R gives

R = ( 5.97 × 10 24 kg ) ( 0 m ) + ( 7.36 × 10 22 kg ) ( 3.82 × 10 8 m ) 5.98 × 10 24 kg + 7.36 × 10 22 kg = 4.64 × 10 6 m.

Significance

The radius of Earth is 6.37 × 10 6 m , so the center of mass of the Earth-moon system is (6.37 − 4.64) × 10 6 m = 1.73 × 10 6 m = 1730 km (roughly 1080 miles) below the surface of Earth. The location of the center of mass is shown (not to scale).

The earth is drawn entered on the origin of an x y coordinate system. The moon is located to the right of the earth on the x axis. R c m is a horizontal vector from the origin pointing to the right, smaller than the radius of the earth.
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Check Your Understanding Suppose we included the sun in the system. Approximately where would the center of mass of the Earth-moon-sun system be located? (Feel free to actually calculate it.)

The average radius of Earth’s orbit around the Sun is 1.496 × 10 9 m . Taking the Sun to be the origin, and noting that the mass of the Sun is approximately the same as the masses of the Sun, Earth, and Moon combined, the center of mass of the Earth + Moon system and the Sun is
R CM = m Sun R Sun + m em R em m Sun = ( 1.989 × 10 30 kg ) ( 0 ) + ( 5.97 × 10 24 kg + 7.36 × 10 22 kg ) ( 1.496 × 10 9 m ) 1.989 × 10 30 kg = 4.6 km
Thus, the center of mass of the Sun, Earth, Moon system is 4.6 km from the center of the Sun.

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Center of mass of a salt crystal

[link] shows a single crystal of sodium chloride—ordinary table salt. The sodium and chloride ions form a single unit, NaCl. When multiple NaCl units group together, they form a cubic lattice. The smallest possible cube (called the unit cell ) consists of four sodium ions and four chloride ions, alternating. The length of one edge of this cube (i.e., the bond length) is 2.36 × 10 −10 m . Find the location of the center of mass of the unit cell. Specify it either by its coordinates ( r CM, x , r CM, y , r CM, z ) , or by r CM and two angles.

The sodium chloride crystal structure is a square lattice, with alternating Sodium (represented as larger green spheres) and Chlorine (represented as smaller red spheres) ions at the intersections. A unit cell is identified as one of the cubes making up the lattice.
A drawing of a sodium chloride (NaCl) crystal.

Strategy

We can look up all the ion masses. If we impose a coordinate system on the unit cell, this will give us the positions of the ions. We can then apply [link] , [link] , and [link] (along with the Pythagorean theorem).

Solution

Define the origin to be at the location of the chloride ion at the bottom left of the unit cell. [link] shows the coordinate system.

An illustration of a unit cell of an N a C l crystal as a cube with ions at each corner. Four green ions are shown and labeled as m 1 at the origin, m 3 at the corner on the diagonal on the x y plane, m 6 at the corner on the diagonal on the x z plane, and m 8 at the corner on the diagonal on the y z plane. Four red ions are shown and labeled as m 2 on the x axis, m 4 on the y axis, m 5 on the z axis, and m 7 on the remaining corner.
A single unit cell of a NaCl crystal.

There are eight ions in this crystal, so N = 8:

r → CM = 1 M ∑ j = 1 8 m j r → j .

The mass of each of the chloride ions is

35.453 u × 1.660 × 10 −27 kg u = 5.885 × 10 −26 kg

so we have

m 1 = m 3 = m 6 = m 8 = 5.885 × 10 −26 kg .

For the sodium ions,

m 2 = m 4 = m 5 = m 7 = 3.816 × 10 −26 kg .

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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