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November 29, 2014

November 29, 2014

Posted by **Michael** on Sunday, December 7, 2008 at 3:21pm.

- astronomy/physics -
**tchrwill**, Sunday, December 7, 2008 at 6:37pmThe gravitational force of attraction of the Sun on the Earth, or the Earth on the Sun, is given by Newtons Universal Law of Gravitation

F = GMm/r^2

where F is the attractive force, G is the Universal Gravitational Constant, 1.069304x10^-9, M is the mass of the Sun, m is the mass of the earth, and r is the distance between them. This force of attraction of the Sun on the earth is what causes the centripetal acceleration which constantly accelerates the Earth toward the Sun, keeping it in its orbit at a mean distance of 92,960,242 miles. This centripetal force is expressed by F = mV^2/r where m is the mass of the earth, V is the orbital velocity of the Earth, and r the radial distance of the Earth from the Sun.

Since these two forces are equal to one another, we may write

F = GMm/r^2 = mV^2/r

from which we derive the mass of the Sun as

M(S) = rV^2/G.

Taking the mean distance between the Earth and the Sun as 92,960,242 miles, and the Earth's mean orbital velocity of 07,741 ft/sec., we can write for the mass of the Sun

M(S) = [(92,960,242(5280))97,741^2]/1.069304x10^-9 = 4.385x10^30 lb. mass.

Applying the same logic to the known information about the Earth and Moon:

The Moons mean orbital velocity is ~3340 ft/sec. while its mean distance may be taken as ~238,868 miles. Therefore, for the mass of the Earth we get

m(E) = [(238,868(5280))3340^2]/1.069304x10^-9 = 1.315x10^30 lbs. mass.

Now apply the same logic to your specific problem.

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