Kepler\'s (Third) Law for the relation between the radius, a, of the circular or

Question

Kepler's (Third) Law for the relation between the radius, a, of the circular orbit of an object around an object of mass M under the influence of the gravitational force between them, and the time it takes to complete a full orbit, P, is a^3 = GM/4 phi^2 P^2 This assumes that the object in orbit has a mass much smaller than M (as is the case for the planets moving around the Sun, or satellites around the 11 Earth). Here, G = 6.67 times 10^-11 m^3 kg^-1 s^-1 is the Gravitational Constant, which comes from Newton's Law of Gravitation. This law should also apply to the orbit of the Sun around the center of the Galaxy-so we can measure the mass of the Galaxy if we know the size of the Sun's orbit, and its period of revolution! Painstaking observations of the distribution and motion of the stars surrounding us in the Galaxy have provided an estimate for the distance to the Galactic Center, a: a = 8.5 kpc (kilo-parsec) = 8.5 times 10^3 times 3.1 times 10^16 m 2.6 times 10^20 m. The orbit of the Sun in the Galaxy is approximately a circle, so a is the radius of that circle. The velocity, v, of the Sun on its orbit is approximately 220 km/sec. How long does it take the Sun to complete one full orbit? Call that period P. The Galaxy is obviously not a single massive thing sitting at the Center, that we orbit around, like a planet orbits the Sun. Instead, the Galaxy is sort of pancake-like. But it is OK for a first guess using Kepler's (Third) Law given above, to assume that most of the mass of the Galaxy is in fact concentrated near the center: that is where most of the stars are (outside the Sun's orbit, things get pretty sparse pretty fast). Use Kepler's Third Law to estimate the mass of the Galaxy from the properties of the Sun's orbit. Convenient units: in, you found and answer for the mass in kilograms. The number you got, on its face, does not have much meaning. It acquires much more meaning if we express the mass in units of the mass of the Sun: M = 2 times 10^30 kg. How big is the units of the mass of the Sun: M estimated mass of the Galaxy in units of the mass of the Sun? And if we assume that the average mass of a star in the Galaxy is about equal to the mass of the Sun (close enough for now), how many stars does the Galaxy have?

Explanation / Answer

part a:

a=2.6*10^20 m

v=220000 m/s

time period P=2*pi*a/v

=2*pi*2.6*10^20/(220000)=7.4256*10^15 seconds

part b:

M=4*pi^2*a^3/(G*P^2)

=4*pi^2*(2.6*10^20)^3/(6.67*10^(-11)*(7.4256*10^15)^2)=1.8866*10^41 kg

part c:

in terms of mass of the sun,

mass of galaxy=1.8866*10^41/(2*10^30)=9.4332*10^10

so there are 9.4332*10^10 strs in the galaxy.

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