<p>You need to use the generalized (a la Newton) form of Kepler's third law; see <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/orbv.html#bo" rel="nofollow">here</a>.</p>

CC BY-SA 3.0
2014-02-14T03:32:35.110Z
The equation M1 + M2 = a^3 / p^2 should solve this, correct? Or am I mistaken?
user1026
CC BY-SA 3.0
2014-02-14T03:38:46.613Z
Yes, but be careful of the units!! Note that equation has the masses in solar masses, the time in years, and the radius in AU.
Scott Griffiths
CC BY-SA 3.0
2014-02-14T04:02:48.100Z
I got 3.71885*10^16 (half of 7.4367*10^16 since it is asking for M1 = M2 = X). Mastering Astronomy says I am incorrect though, do you know where I messed up? I converted ten months to seconds (2.628*10^7) since P is in meters/second. I got 3.3024*10^12 for A also. My equation was ((4π^2)/(6.67*10^-11))*((3.3024*10^12)/(2.628*10^7)). Any idea what I did wrong...?
user1026
CC BY-SA 3.0
2014-02-14T04:06:10.907Z
Just realized I forgot to divide my equation ((4π^2)/(6.67*10^-11))*((3.3024*10^12)/(2.628*10^7)) by the Sun's mass (2*10^30). The answer is still coming out wrong though... now I got 1.859*10^-14. Ahh :(
user1026
CC BY-SA 3.0
2014-02-14T04:36:39.087Z
Fixed some errors, still coming out incorrect though...  ((4π^2)/(6.67*10^-11))*(((3.3460735*10^11)^3)/((2.628*10^7)^2))
user1026
CC BY-SA 3.0
2014-02-15T01:39:03.157Z
Anyone? :( I'm at a loss and my Professor hasn't responded to my emails
user1026
CC BY-SA 3.0
2014-02-14T03:06:21.707Z
2014-02-14T03:06:21.707Z
Scott Griffiths
<p>We can assume that the stars are equal in mass, and their orbits are circular</p>

<p>The orbital speed is 80000 m/s and at an orbital period of 10 months (or $2.628\times 10^7$s) the length of the orbit is $2.1024\times 10^{12}$ m or 14.05 AU The radius of the orbit therefore is $14.05/\tau$ = 2.237AU. </p>

<p>The version of Keplers law given is $$T^2 =\frac{a^3}{m_1+m_2}$$</p>

<p>substituting $T^2 =(10/12)^2 = 0.6944$ (divide by 12 to convert to years)  and $a^3= 11.19$ gives $$m_1+m_2 = \frac{11.19}{0.6944} = 16.12\  \mathrm{solar\ mass}.$$ </p>

<p>Since $m_1=m_2$, the mass of each star is 8.06 solar masses, or $1.6\times 10^{31}$kg</p>

CC BY-SA 3.0
2016-04-05T16:06:36.800Z
2016-04-05T16:06:36.800Z
James K
<p><em>If anyone would at least be willing to point me in the direction of finding the answer, or help me solve it I'd appreciate it. I don't know how to go about this, and need to know how to solve this on my own (plus I need the answer for a homework assignment, but just an answer isn't gonna help).</em></p>

<p><em>Question:</em></p>

<p>The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 10 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of $~8.0\cdot 10^4~ \mbox{m}/\mbox{s}.$ 
What are the masses of the two stars? Assume that each of the two stars traces a circular orbit around their center of mass.</p>

<mass>
What are the masses of the two stars (given the information provided)?
user1026
# What are the masses of the two stars (given the information provided)?

(Asked by: username_0 on 2014-02-14)

*If anyone would at least be willing to point me in the direction of finding the answer, or help me solve it I'd appreciate it. I don't know how to go about this, and need to know how to solve this on my own (plus I need the answer for a homework assignment, but just an answer isn't gonna help).*

*Question:*

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 10 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of $~8.0\cdot 10^4~ \mbox{m}/\mbox{s}.$
What are the masses of the two stars? Assume that each of the two stars traces a circular orbit around their center of mass.

**Tags:** mass

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## 2 answers

### Answer by username_1 on 2014-02-14. Score: 1

You need to use the generalized (a la Newton) form of Kepler's third law; see [here](http://hyperphysics.phy-astr.gsu.edu/hbase/orbv.html#bo).

> **Comments:**
> **username_0**: The equation M1 + M2 = a^3 / p^2 should solve this, correct? Or am I mistaken?  
> **username_1**: Yes, but be careful of the units!! Note that equation has the masses in solar masses, the time in years, and the radius in AU.  
> **username_0**: I got 3.71885*10^16 (half of 7.4367*10^16 since it is asking for M1 = M2 = X). Mastering Astronomy says I am incorrect though, do you know where I messed up? I converted ten months to seconds (2.628*10^7) since P is in meters/second. I got 3.3024*10^12 for A also. My equation was ((4π^2)/(6.67*10^-11))*((3.3024*10^12)/(2.628*10^7)). Any idea what I did wrong...?  
> **username_0**: Just realized I forgot to divide my equation ((4π^2)/(6.67*10^-11))*((3.3024*10^12)/(2.628*10^7)) by the Sun's mass (2*10^30). The answer is still coming out wrong though... now I got 1.859*10^-14. Ahh :(  
> **username_0**: Fixed some errors, still coming out incorrect though...  ((4π^2)/(6.67*10^-11))*(((3.3460735*10^11)^3)/((2.628*10^7)^2))  
> **username_0**: Anyone? :( I'm at a loss and my Professor hasn't responded to my emails  

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### Answer by username_2 on 2016-04-05. Score: 0

We can assume that the stars are equal in mass, and their orbits are circular

The orbital speed is 80000 m/s and at an orbital period of 10 months (or $2.628\times 10^7$s) the length of the orbit is $2.1024\times 10^{12}$ m or 14.05 AU The radius of the orbit therefore is $14.05/\tau$ = 2.237AU.

The version of Keplers law given is $$T^2 =\frac{a^3}{m\_1+m\_2}$$

substituting $T^2 =(10/12)^2 = 0.6944$ (divide by 12 to convert to years) and $a^3= 11.19$ gives $$m\_1+m\_2 = \frac{11.19}{0.6944} = 16.12\ \mathrm{solar\ mass}.$$

Since $m\_1=m\_2$, the mass of each star is 8.06 solar masses, or $1.6\times 10^{31}$kg


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