21st Century Astronomy (Full Fourth Edition) by Laura Kay

Author:Laura Kay
Language: eng
Format: mobi
Published: 0101-01-01T00:00:00+00:00

5. The stars are in circular orbits. You know this because

G( m + m )

1

2

the Doppler velocities about the star are symmetric; the

Rearranging this equation a bit turns it into an expression for

approach and recession speeds of the star are equal.

the sum of the masses of the two objects:

These data are summarized in Figure 13.15. You begin

A 3

m + m = 4π2

____ ×

___

your analysis by noting that the star is an eclipsing binary,

1

2

G

P 2

which tells you that the orbit of the star is edge-on to your

We can ignore the value of 4π2/ G by applying what we know

line of sight. The Doppler velocities tell you the total orbital

about Earth’s orbit around the Sun, and then expressing m, A,

velocity of each star, and you determine the size of the orbits

and P as ratios. If A = 1 AU and P = 1 year, M = mass of the

£

using the relationship Distance = Speed × Time. In one or-

Sun and M = mass of the Earth, then we know that m + m =

¢

1

2

bital period, star 1 travels around a circle—a distance of

M + M ≈ M (because M is much larger than M ). So if we

£

¢

£

£

¢

express the masses m and m , A, and P in that equation in terms

d = (20.4 km/s) × (2.63 yr) = 53.7 km × yr/s

1

2

of Solar System units—such as m = m /(1 M ), A = A/(1 AU),

1

1

£

AU

and P

= P/(1 year)—then this equation simplifies to:

You multiply by the number of seconds in a year:

years

m

m

( A ) 3

km × yr

3.16 × 107 s

d = 53.7

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