Astronomy C Question Marathon

[alpacas]

Looking back actually I'm not sure (but trying to figure it out that number must have come from somewhere) because distance modulus suggests that it would be

D~2000 LY=613.5 pc
and with m=16 (http://en.wikipedia.org/wiki/NGC_3132)
absolute magnitude would be 7, which is obviously really different

also AlphaTauri the GRS1915+105 lower mass limit calculation is explained here http://arxiv.org/pdf/astro-ph/0111540.p ... ion_detail, also explained here http://works.bepress.com/cgi/viewconten ... ic_addison

[XJcwolfyX]

Shall we keep this going? Anyone want to ask a question?

[Crazy Puny Man]

How about this?

Taken directly from one of my invite tests:

A binary star system at an inclination 10 degrees from our view line is noticed to have a separation of .5 AU and a observed value of 4016.23 angstroms for a wavelength that should be at 4000 angstroms from one star and an observed value of 3508.52 angstroms for a wavelength that should be at 3500 angstroms from the other star. Please compute:
a. The distance to the center of mass point from the larger star (meters)
b. The period of the system (seconds)
c. The mass of the larger star (solar masses)

[cellobix]

I noticed that both stars were redshifted. I'm not sure how to apply Hubble's Law to remedy that in this situation (i.e. how to calculate a distance from the given information), so I assumed one of the stars was supposed to be blueshifted.

Solution - I feel my answer to part C is ridiculously large, but I can't spot my mistake (Click to Show Hidden Text)

Let Star A be shifted from 4000A to 4016.23A.
Let Star B be shifted from 3500A to 3508.52A.

Using redshift and the Doppler Effect, calculate the radial velocities.
rad. vel. A = 1 217 000 m/s
rad. vel. B = 730 300 m/s

Calculate the speed using basic right triangle trig.
speed A = 1 236 000 m/s
speed B = 741 600 m/s

Since star B is slower, it is the more massive.  Use d=rt to conclude that the distance from the center of mass is proportional to orbital speed, thus:
(dist B) / (dist A) = (speed B) / (speed A) = 1.667
EDIT: this is backwards.  The ratio in this case is 1/1.667.  Oops...

It is given that the stars are separated by .5 AU, thus:
(dist B) + (dist A) = .5 AU = 7.48 E+10 m.

Solve this system for (dist B) to obtain a distance of 2.80 E+10 m.

Next, use d=rt to obtain an orbital period of 238 000 seconds.

Finally, convert a unit and apply Kepler's third law:
238 000 seconds = .00753 years
(total mass) = (7.48 E+10 m)^3 / (.00753 years)^2 = 2204 solar masses.

Apply the ratio (mass A) * (dist A) = (mass B) * (dist B) to obtain a (mass B) of 1380 solar masses.

Summary:
a. 2.80 E+10 m
b. 238 000 seconds
c. 1380 solar masses

[Crazy Puny Man]

I noticed that both stars were redshifted. I'm not sure how to apply Hubble's Law to remedy that in this situation (i.e. how to calculate a distance from the given information), so I assumed one of the stars was supposed to be blueshifted.

Solution - I feel my answer to part C is ridiculously large, but I can't spot my mistake (Click to Show Hidden Text)

Let Star A be shifted from 4000A to 4016.23A.
Let Star B be shifted from 3500A to 3508.52A.

Using redshift and the Doppler Effect, calculate the radial velocities.
rad. vel. A = 1 217 000 m/s
rad. vel. B = 730 300 m/s

Calculate the speed using basic right triangle trig.
speed A = 1 236 000 m/s
speed B = 741 600 m/s

Since star B is slower, it is the more massive.  Use d=rt to conclude that the distance from the center of mass is proportional to orbital speed, thus:
(dist B) / (dist A) = (speed B) / (speed A) = 1.667

It is given that the stars are separated by .5 AU, thus:
(dist B) + (dist A) = .5 AU = 7.48 E+10 m.

Solve this system for (dist B) to obtain a distance of 2.80 E+10 m.

Next, use d=rt to obtain an orbital period of 238 000 seconds.

Finally, convert a unit and apply Kepler's third law:
238 000 seconds = .00753 years
(total mass) = (7.48 E+10 m)^3 / (.00753 years)^2 = 2204 solar masses.

Apply the ratio (mass A) * (dist A) = (mass B) * (dist B) to obtain a (mass B) of 1380 solar masses.

Summary:
a. 2.80 E+10 m
b. 238 000 seconds
c. 1380 solar masses

You can assume the redshifts were calculated at different times in the star's orbit.

Your answers are correct Um, B is indeed closer to the star, but your ratio (dist B) / (dist A) = (speed B) / (speed A) = 1.667 implies that B is farther/faster...

[cellobix]

The orbit of an eclipsing binary system is oriented such that it is observed from directly edge-on. The system has a period of 68 hours, the secondary star has an orbital velocity of 72.5 km/s, and the eclipse of the primary star lasts 11 hours. What is the diameter of the primary star?

[Crazy Puny Man]

The orbit of an eclipsing binary system is oriented such that it is observed from directly edge-on. The system has a period of 68 hours, the secondary star has an orbital velocity of 72.5 km/s, and the eclipse of the primary star lasts 11 hours. What is the diameter of the primary star?

Is it just (Click to Show Hidden Text)

72.5 km/s * 11 hr * 3600 s/hr = 2.87 x 10^6 km?

[cellobix]

Nope.

[Crazy Puny Man]

Nope.

When you say "the eclipse of the primary," are we talking primary minimum, or when the secondary star eclipses the primary star?

Also, on an eclipsing binary light curve, is this 11 hours = T3 - T1? Or T4 - T1?

[cellobix]

When you say "the eclipse of the primary," are we talking primary minimum, or when the secondary star eclipses the primary star?

I am referring to the secondary star eclipsing the primary.

Also, on an eclipsing binary light curve, is this 11 hours = T3 - T1? Or T4 - T1?

I'm not sure what you mean by T1, T3, and T4, sorry.