The Physics Factbook

Edited by Glenn Elert -- Written by his students

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

Beichner, Robert J., John W. Jewett, and Raymond A. Serway. Physics For Scientists and Engineers. New York: Saunders College, 2000. |
"Body: Earth, Mass (kg): 5.98 × 10^{24} kg" |
5.98 × 10^{24} kg |

"Earth." The World Book Encyclopedia. Vol. 6. Chiacago: World Book Inc., 2001. |
"Mass: 6,600,000,000,000,000,000,000 (6.6 sextillion) short tons (6.0 sextillion metric tons)." | 6.0 × 10^{24} kg |

Smith, Peter J. The Earth. New York: Macmillan Company, 1986. |
"The Solar System is dominated by the Sun, which has a mass of about 2 × 10^{30} kg. This is about 343,000 times the mass of the Earth." |
5.83 × 10^{24} kg |

Larouse. Astronomy. New York: Facts on File Publications, 1981. |
"The calculation shows that it is very close to 6.10^{24} kilograms (6,000 billion billion tons)." |
6.0 × 10^{24} kg |

As we all learned in grade school, the Earth is the third planet from the
Sun. The planet Earth is only a tiny part of the universe, but it is the
home of human beings and all known life in it. Animals, plants and other
organisms live almost everywhere on Earth's surface. It ranks fifth in
size, and its mass is found to be about 5.98 × 10^{24} kg.
Mass is a characteristic that is inherent, and it is independent of the
object's environment and the method used to measure it. It is a scalar
quantity, that is, a single value with and appropriate unit that has no
direction.

The mass of the Earth may be determined using Newton's law of gravitation.
It is given as the force (F), which is equal to the Gravitational constant
multiplied by the mass of the planet and the mass of the object, divided
by the square of the radius of the planet. We set this equal to the fundamental
equation, force (F) equals mass (m) multiplied by acceleration (a). We
know that the acceleration due to gravity is equal to 9.8 m/s^{2},
the Gravitational constant (G) is 6.673 × 10^{-11} Nm^{2}/kg^{2},
the radius of the Earth is 6.37 × 10^{6} m,
and mass cancels out. When we rearrange the equation and plug all the
numbers in, we find that the mass of the Earth is 5.96 × 10^{24} kg.

F = Gm_{1}m_{2}/r^{2} = ma

Gm/r^{2} = g

m = gr^{2}/G

m = (9.8 m/s^{2})(6.37 × 10^{6} m)^{2}/(6.673 × 10^{-11} Nm^{2}/kg^{2})

m = 5.96 × 10^{24} kg

The Earth gains mass each day, as a result of incoming debris from space. This occurs in the forms of "falling stars", or meteors, on a dark night. The actual amount of added material depends on each study, though it is estimated that 10 to the 8th power kilograms of in-falling matter accumulates every day. The seemingly large amount, however, is insignificant to the Earth's total mass. The Earth adds an estimated one quadrillionth of one percent to its weight each day.

Samantha Dong -- 2002

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

Hewitt, Paul G. Conceptual Physics. N.P.: Addison-Wesley Publishing, 1987: 147. |
"M = 5.98 × 10^{24} kg" |
5.98 × 10^{24} kg |

Characteristics of Earth. Encarta. 5 May 2002. | "Approximate Mass: 5.98 × 10 ^{21} metric tons" |
5.98 × 10^{24} kg |

Arnett, Bill. Earth. 05 May 2002. 27 May 2002. | "mass: 5.5.972e24kg" | 5.972 × 10^{24} kg |

Giancoli, Douglas C. Physics. Englewood Cliffs, NJ: Prentice-Hall, 1980. 73. |
"m_{e} = gr^{2}/G = … = 6.0 × 10^{24} kg" |
6.0 × 10^{24} kg |

The Earth. Enchanted Learning. 1999. | "The Earth's mass is about 5.98 × 10^{24} kg" |
5.98 × 10^{24} kg |

Earth, the third planet from the sun, is one of the most unique celestial bodies in our solar system. It is the only planet in our solar system that sustains life and it is the planet that we can call our own.

In approximately 230 BC, the Greek mathematian, Eratosthenes calulated the
radius of the Earth. He compared the shadows in the wells during the summer solstice
and obtained the value 6.38 × 10^{6} M. In the 16^{th} century, Galileo determined the acceleration due to the force of gravity near
the surface of the Earth and obtained 9.8 m/sec^{2}.

Sir Isaac Newton greatly contributed to the study of physics and therefore, his efforts determined the mass of the Earth. His law of gravity and second law of motion are used together to obtain a value for the mass of our planet. Newton's law of gravity formulates the gravitational force that two masses exert on each other and is given by

F = GmM/r^{2}

M an m are the two masses, r is the separation between them, and G is the universal
gravitational constant which was calculated by Henry Cavendish in 1798, which
has a value of 6.67 × 10^{-11} m^{3}/(kg sec^{2}).

If we assumed that M is the mass of the Earth, and m is the mass of an object on the surface of the Earth, we can solve for M by equating Newton's Law of Gravity with his second law of motion

F = ma

We have:

F = GmM/r^{2} = ma → GM/r^{2} = a

Solving for M, the mass of the Earth, and using

Where

a = 9.8 m/s^{2},

r = 6.38 × 10^{6} m, and

G = 6.67 × 10^{-11} m^{3}/(kg sec^{2})

we obtain:

M = ar^{2}/G = 5.98 × 10^{24} kg.

Christine Lee -- 2002

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