What was 29(a)...can't remember them at all.23 (b) the orbital velocity, what did you guys get and how?
and also for 29 (a)?
Fuck lol I miscalculated 29(a)...sighI think it was about 7692ms-1?
29(a) would be 2.26x10^15 photons
That is correct.I think it was about 7692ms-1?
29(a) would be 2.26x10^15 photons
The standard orbital velocity equation, v=ROOT(Gm/r).What was the method of working out the orbital velocity question. I can't remember my answer but if i see the method i will know if i got it right or not because i remember how i did it. I can hardly remember anything from part B.
I think i found what the period was by r^3/T^2 = GM/4pi^2 and then solved v with V = 2rpi/t Is this wrong because i assumed its orbit as a circle? I think i overcomplicated this one, easy marks lost if i did.The standard orbital velocity equation, v=ROOT(Gm/r).
You were given the Earth's radius and the altitude of the satellite, and since you know G, then solve for V.
You always assume it's a circle since all of the equations contain the radius. v=[2(pi)r]/T probably would have worked. I think the values were 350KM altitude, 6730KM Earth radius and 5.97x10^24kg Earth mass. Try both methods and if you get the same answer, you can probably expect full marks. Although, the marker would probably be shaking his/her headI think i found what the period was by r^3/T^2 = GM/4pi^2 and then solved v with V = 2rpi/t Is this wrong because i assumed its orbit as a circle? I think i overcomplicated this one, easy marks lost if i did.
yep it got the right answer. Silly of me going the long way round though hah. I'm just relieved i didn't lose a couple easy marks.You always assume it's a circle since all of the equations contain the radius. v=[2(pi)r]/T probably would have worked. I think the values were 350KM altitude, 6730KM Earth radius and 5.97x10^24kg Earth mass. Try both methods and if you get the same answer, you can probably expect full marks. Although, the marker would probably be shaking his/her head
Goodieyep it got the right answer. Silly of me going the long way round though hah. I'm just relieved i didn't lose a couple easy marks.
Do you reckon that'll be minus 1/2/3 ?Dammit i forgot the radius of the earth
I guess -1 mark?Do you reckon that'll be minus 1/2/3 ?
The answer was 7692m/s. Someone put up what the radius and mass was earlier in this thread. Find the values in this thread and go back through the method and see if you got the right answer. If you got the right answer then i don't see why you wouldn't get full marks.Could some tell me if what I did was right..
I made F=mv2/r and F=ma become mv2/r=ma, then I found the value of a at the height of the rocket (a=g) using g=Gm/r2 which gave me an answer of around 8.7... I then rearraged the formula and eventual got to v=ROOTar.. Got an answer somewhere in the seven thousands, not sure of the exact value, but I have a feeling I probably fluked it.. Could someone please tell me if what I did could get me any marks ?
I used this as well and got the right answer. The answer was posted earlier in the thread, it was somewhere in the 7000s.i got about 7000m/s, using the GM/4piesquared = r^3 / T^2 and v = 2pieR/T method. Your method works i think as well
i got this aswell..The answer was 7692m/s. Someone put up what the radius and mass was earlier in this thread. Find the values in this thread and go back through the method and see if you got the right answer. If you got the right answer then i don't see why you wouldn't get full marks.
yeah i got this too xD but i was being careful with the significant figures and stuff cause considering this paper being too easy>.. markers can be mean and mark on significant figuresI think it was about 7692ms-1?
29(a) would be 2.26x10^15 photons