Official OCR A2 G484 June 2016 [Module 1] [90UMS]

Well, they told you the period was equal for both of them, 4 days.
I've no idea if that is right.
I just used keplers third law with T^2 = (4Pi^2 * R2^3)/(G*M2)
That got me 1.73*10^33.
Was it out of 2 or 3?

i did this very differently

Got

F=mv^2/r equal to the first eqn we wrote F=Gm1m2/(r1+r2)^2

I had V for planet 1 and R1 so put equal to each other M1 cancels in both equations

Leaving

V^2/r1 = Gm2/(r1+r2)^2 leaving only M2 as a variable,

Not sure if anyone else did it it like this.

Got 1x10^30 something like that anyway



Starting to forget answers but SHC of the metal was 128.something ???

Yeah a couple people in my class got that too. I've done it like 10 times and I consistently get 10^33, perhaps you mis-entered something on your calculator?

Workings to get the mass of S2 for those wondering.

Workings to get the mass of S2 for those wondering.

Are you sure this is right I don't really follow what you did

You know that the gravitational force exerted on the star is equal to the resultant force. Therefore the gravitational force exerted on the star = centripetal force. You know S1 is 3 x greater mass than S2 so you can sub in S1 = 3S2 into both sides of the equation. From this point you can solve for S2

I got 130.8 for it

makes sense its just that I was one of the people who got 1.73x10^33, any ideas what those people(me) did wrong? or you might have done wrong? I honestly have no clue, besides I thought the period was years?

You know that the gravitational force exerted on the star is equal to the resultant force. Therefore the gravitational force exerted on the star = centripetal force. You know S1 is 3 x greater mass than S2 so you can sub in S1 = 3S2 into both sides of the equation. From this point you can solve for S2

I shall not believe you until I see TeacherCol's solutions. You're method is FAR too complicated for 3 marks and is based on so many assumptions and what you can imply from the data. This is not astrophysics, it's plain simple clear cut Newtonian World Physics. So many people here got answers in the order of 10^33 and not one has got your answer in the order of 10^35. Clearly something is wrong right?

just need to say one thing, the period was 4 years not days as everyone seems to be saying, I worked it out and if it was days both stars would have to be travelling above the speed of light (clearly impossible)

I shall not believe you until I see TeacherCol's solutions. You're method is FAR too complicated for 3 marks and is based on so many assumptions and what you can imply from the data. This is not astrophysics, it's plain simple clear cut Newtonian World Physics. So many people here got answers in the order of 10^33 and not one has got your answer in the order of 10^35. Clearly something is wrong right?

There was also 1 mark for letters question, and 2 for the one below that, and 3 for the one below that (just before the radius ratio question)

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I am 100% sure that this person Is correct. I read it very carefully. You were supposed to get that equation shown above and that raito, what was it 3R1=R2 or something, then solve simultaneously. it came out with R1=1.2x10^12 and R2=3.6x10^12.