Binary Star Systems

[@MarinaGattas]

I just wanted to say thank you for posting all those videos. I live in monterey, ca and I can honestly say I wouldnt have made it through my physics class without your help! thanks again!!

[@ayushrajmehrotra1318]

This video is simply excellent. To the point and thats just everything I need to know about binary stars. Thanks a lot.

[@RajKumarGupta-yz4ye]

I am an Indian... And I liked your video... Keep doing ...

[@xLouDNoiSeSx]

You should make some videos for Modern Physics and also Classical mechanics as those are on my calender for the year.

[@jimdogma1537]

All these videos are really good, thanks for posting them. Hopefully, you will make more in the future, perhaps some relativity and modern physics. I have some minor comments on this series, although stating them isn't going to do much good since the series has already been completed 2 years prior. However, if you make more videos, please try to follow through a problem till the end so we can check our results. You often seem to bail on the solution 2 or 3 steps before it's finished.

[@tilakrajnarang5896]

Will you plz elaborate on the reduced mass approach

[@thelazyapple9994]

😭😭 Thank you so much...what would have I done without you!

[@lotusnaturals1897]

when you consider the masses of the stars, do you mean compared to our star (like 1=our star) or like grams? what does G represent? do the r1/r2 distances have to be in AU or km or something else?

[@tianchenzhang470]

G is a constant which is 6.67*10^-11hopefully you have already known that

[@JohnFisher55]

Would this every explain whether or not the solar system nearly became a binary star system instead of a single star with planets. We may speculate that in that case the Earth might not even exist, or even if it existed would be in an orbit giving surface conditions. Can a binary star system be possible where in one star is stationary and the other star revolves around it? Is the sun part of the bsystem?

[@23surrea1]

can you explain why for the second mass, the radius for Fnet is r1+r2? Shouldn't it just be r2?

[@ceulgai2817]

For Newton's gravitational formula, you use the distance the bodies' centers of mass; hence, r1+r2.

[@devadharsan7721]

sir this video was so clear and easy to understand thank you for this

[@seraphinn1000]

Thank you for help - very helpful.... just wondering how you find the inclination to do a geometrical correction to the radial velocities in order to find the true orbital velocities?

[@artisticnoob5003]

Short and precise! Loved it :)

[@akashadak4160]

Thank you so much!! I have a question.... will they have same time period?? How??

[@lasseviren1]

Yes, the two stars will take the same time to go around their shared center of mass because they will always be on opposite sides of this center of mass. Sort of like to dancers holding hands and rotating around their combined center of mass.

[@akashadak4160]

@@lasseviren1 okay sir Thank you so much

[@shideshpatil4796]

Why they not collide?

[@lasseviren1]

If you took away their relative velocities, they would collide.

[@davepastern]

would not the radial velocity around the common mass point for m2 be faster than that of m1? I struggle to see why m1 and m2 would have the same orbital period, given the larger orbit of m1 in your example. Note: it has been many years since I did high school level physics, and nothing as complex as this (basic forces/motion and some vector work).

[@lasseviren1]

The two stars will have the same period about their common center of mass because they will always be on opposite sides of the center of mass. In the video's example, m1 (the less massive star) does have a greater speed but it also has a larger orbital radius, which allows for the two periods to be equal. Think about two different-sized skaters spinning in a circle around a common center. They have to always be opposite to each other. Finally, I wish you all the best in your renewed quest to study more physics.

[@davepastern]

@@lasseviren1 ah yes, that makes sense. As to studies, nah, not at my age. My brain has slowed down the pst few years and the maths is now beyond me sadly.. I can learn it, but not at the pace that modern universities push. They seem to design courses to suit 18 year olds, and for older people, it's tough luck.

[@fantom1086]

What if thier com moved

[@LoL-bk4kw]

Too old video but right on point ☝🏿

[@tianacole6719]

Thank you so much this video was so helpful!

[@krishanginanda2876]

Thankyou so much! It really helped me 😄

[@bkboggy]

Isn't it supposed to be mv^2 / r , not v^2 / r (centripetal force)? The circular orbit equation comes out to be v = sqrt(Gm/r)

[@ardent146]

Fnet=ma=mv^2/r So, a=v^2/r

[@xLouDNoiSeSx]

Thanks for all the videos!

[@chethans5908]

Thank you professor

[@dimplebabbar5945]

This question was asked in bitsat 2k18😨

[@ayoopdog]

very fascinating. Thanks for uploading!

[@surendrakverma555]

Very good lecture

[@kalinagoheyokaempathyindig76]

I discovered are sun has a binary star system it is here now close proximity to are sun ive been tracking it for a year now!

[@hannonbeau]

I do thank you for time, however there is no doubt in my mind Sol has a binary twin, hence we live in a binary solar system, your rotational models are incorrect.

[@subhrasnatachakraborty8311]

Please correct ur mistake ...... U r taking a=v*v/r its completely wrong as we know a= v*v/2r so........ Please don't misguide students

[@lasseviren1]

What?

[@subhrasnatachakraborty8311]

lasseviren1 its wrong

[@Danny-sj6es]

The video is correct. We are considering the acceleration of just one of the stars, but either way what your proposing doesn't make sense. The V^2 / r is derived from circular motion look it up and then re watch the video.

[@dimplebabbar5945]

India ki besti karara

[@epidemicrage4337]

Equate centripetal force with Gravitational force. It’s the same equation