# Density of the Earth

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Bibliographic Entry | Result (w/surrounding text) |
Standardized Result |
---|---|---|

Neff, Robert F. & Zitewitz, Paul W. Physics, Principles and Problems. New York: Glencoe, 1995: 159. |
"Mass of the Earth 5.979 × 10^{24} kgRadius of the Earth 6.3713 × 10 ^{3} km" |
5.519 g/cm^{3} |

Compton's Interactive Encyclopedia. Compton's, 1995. |
"They divide the mass of the Earth by the volume, which gives the average density of the material in the earth as 3.2 ounces per cubic inch (5.5 g/cm^{3})." |
5.5 g/cm^{3} |

Orbits Voyage Through The Solar System. Phoenix, AZ: Software Marketing, 1989. |
"Mean Density: (water = 1) 5.52." | 5.52 g/cm^{3} |

Morse, Joseph Laffan. Funk & Wagnalls Standard Reference Encyclopedia. New York: Standard Reference Works, 1967: 2934. |
"The average density of the planet [Earth] is 5.52" | 5.52 g/cm^{3} |

Hamilton, Calvin J. Earth Introduction. Views of the Solar System. | "Mean density (g/cm^{3}) 5.515." |
5.515 g/cm^{3} |

The density of the Earth is higher than that of any other planet in our solar
system. Sources vary when it comes to the density of the Earth. All the numbers
that were provided are so close to each other, however, that they can each
be considered valid. Some assorted numbers given would be: 5.5, 5.52, and 5.15 g/cm^{3} (estimations
that are made can change the outcome of a calculation).

Density is found by dividing the mass by the volume (ρ = *m*/*V*).
A scientist named Henry Cavendish is known for calculating the mass (and then
density) of the Earth. Cavendish assembled an apparatus that consisted of a
suspended metal rod with two lead balls hanging from it. He placed masses of
metal near these balls in order to measure the force of attraction between
them. Correspondingly, he could then find the attraction on a mass the size
of the Earth and then determine its density. This famous procedure is known
as the Cavendish Experiment.

In order to find the volume of the Earth you need more information than just
the volume of a sphere formula. This formula (4/3πr^{3}) requires
the radius of the Earth. The diameter of the Earth at the equator is 7926.68 miles
(or 12756.75 km). Now, to find the radius, divide the diameter by 2 (because
any radius is exactly half of its diameter).

The mass of the Earth is found to be 6 sextillion, 587 quintillion short
tons (or 5.98 × 10^{21} metric tons). Since the Earth is
a sphere, the formula 4/3πr^{3} is used to find the volume. The
volume of the Earth is considered to be
1.08 × 10^{12} km^{3} (or 2.5988 × 10^{11} miles^{3}).

I also calculated numbers for the density of the Earth. I knew the mass of
the Earth in grams is 5.979 × 10^{27}. Also, the
radius in centimeters is 6.3713 × 10^{8}. I plugged
this information into the 4/3πr^{3} formula in order to find
the volume. I got the answer 1.084227366 × 10^{27} cm^{3}.
I then divided the volume into the mass and got 5.519 g/cm^{3} as
the density of the Earth.

Since the Earth's mass is so great, it makes a gravity that compresses it more than every other kind of inner planet. The Earth actually lacks all the huge "vapor envelopes" of the gas giants, so almost all of the matter on our planet consists of heavy solids. This is what makes Earth have the highest average density compared to any other planet in the solar system.

Katherine Malfucci -- 2000