Kepler's Third Law

  • Thread starter Morpheus
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  • #1
I'm not good at this stuff, so I need some help from some smarter people...:smile:

The speed of the Sun in its orbit about the Galactic center is 220 km/sec. Its distance from the center is 8500 pc. Assuming that the Sun's orbit is circular, calculate the mass of the Galaxy (in solar masses) interior to the Sun's orbit from Kepler's Third Law. Recall that the circumference of a circle is 2 pi r, where r is the radius. Remember that 1 pc is 206265 AU and 1 AU is 1.5x10^8 km.

Any help would be greatly appreciated!!!
 

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  • #2
sridhar_n
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...

Assuming that the mass of the galaxy is concentrated at its center, the force of the galaxy of mass M on the sun of mass m seperated by a distance r from the galaxy center is:

F = GMm/r^2. This force will be equal to the centrifugal force due the revolution of the sun around the galaxy.
Thus,

GMm/r^2 = mV^2/r, where, V is the velocity of sun around the galaxy.

From this u can calculate the value of M - the mass of the galaxy (Its, just mere substitution...)


Sridhar
 
  • #3
HallsofIvy
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sridhar_n's response is exactly what you need but I am puzzled about the reference to "Kepler's third law".

Kepler's third law is: The ratio of the squares of the revolutionary periods for two planets is equal to the ratio of the cubes of their semimajor axes.

Of course one can extend this to stars orbiting the galaxy but this says nothing about mass. Kepler's laws were purely descriptive. Kepler knew nothing about gravitational force or how mass affected orbital period.
 
  • #4
Thanks for the help. All I know is my Astronomy professor asked us this question and told us to refer to Kepler's Third Law.
 

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