Estimating the mass of our galaxy, and the number of stars in our galaxy

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SUMMARY

The discussion focuses on estimating the mass of the galaxy using Kepler's third law. The formula used is \(((4\pi^2)r^3)/(G \cdot T^2)\), where \(T\) is the period converted to seconds (6.3072 x 1015 seconds) and \(r\) is the radius of 2.85 x 1020 meters. The conversion of light-years to meters is essential for consistency in units, particularly when using the gravitational constant \(G\) in MKS units. Understanding the derivation of this formula is crucial for accurate calculations.

PREREQUISITES
  • Understanding of Kepler's laws of planetary motion
  • Familiarity with the gravitational constant \(G\) in MKS units
  • Ability to convert units, specifically light-years to meters
  • Basic knowledge of astrophysics and celestial mechanics
NEXT STEPS
  • Research the derivation of Kepler's third law in detail
  • Learn about the gravitational constant \(G\) and its applications in astrophysics
  • Study unit conversion techniques, particularly for astronomical distances
  • Explore methods for estimating the mass of celestial bodies
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Astronomers, astrophysicists, students of physics, and anyone interested in calculating the mass of galaxies and understanding celestial mechanics.

aj_17
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Homework Statement
The Sun rotates about the center of the Milky Way Galaxy at a distance of about 30000 light-years from the center
(1ly=9.5×10^15m).
Part A
If it takes about 200 million years to make one rotation, estimate the mass of our Galaxy. Assume that the mass distribution of our Galaxy is concentrated mostly in a central uniform sphere.
Part B
If all the stars had about the mass of our Sun (2×10^30kg), how many stars would there be in our Galaxy?
Relevant Equations
((4pi^2)r^3)/G*T^2)
A previous thread outlined the problem with a correct answer, however I don't understand where they got the formulas from. Here are the steps I've taken so far:
1. Convert 2*10^8 years to seconds = 6.3072*10^15 seconds (period,T)
2. The previous thread then went on to say you plug period and radius (which the thread said was 2.85*10^20, how is this found?) into the "formula":
((4pi^2)r^3)/G*T^2) How do you arrive at this formula?
 
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aj_17 said:
1. Convert 2*10^8 years to seconds = 6.3072*10^15 seconds (period,T)
Looks OK.
2. The previous thread then went on to say you plug period and radius (which the thread said was 2.85*10^20, how is this found?)
You need to get into a consistent set of units. Since you will probably use G in MKS units, you need to convert the radius in light-years into meters.
into the "formula":((4pi^2)r^3)/G*T^2) How do you arrive at this formula?
It's from Kepler's third law.
 

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