Are you going to make an unofficial MS?(Original post by CD223)
Absolute magnitude was reading off of a graph so everyone got slightly different values. Try and plug into your calculator slightly different values (1.981, 1.982, 1.983 etc) and you'll see they vary dramatically.
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AQA Physics PHYA5  Thursday 18th June 2015 [Exam Discussion Thread] watch
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 19062015 20:15

CD223
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 3722
 19062015 20:16
(Original post by ubisoft)
Are you going to make an unofficial MS?
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 3723
 19062015 20:17
(Original post by CD223)
Absolute magnitude was reading off of a graph so everyone got slightly different values. Try and plug into your calculator slightly different values (1.981, 1.982, 1.983 etc) and you'll see they vary dramatically.
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CD223
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 3724
 19062015 20:18
(Original post by ImNormal)
None of them vary as much as his though. His is essentially 9.7x10^6 which is way out. It should be to power of 3
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 3725
 19062015 20:20
Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit2) (i) Six marker, Don't ask.(ii) m  M = 5log25/10 = 0.00973)(i) T = 16111K(ii) The star was actually a white dwarf but I messed up and put main sequence(iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in its n = 2 state and the deep prominent absorption lines in theBalmer series in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s 
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 3726
 19062015 20:25
Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11
(ii) Show that theta = 3.12x10 6 radians
(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)
(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit
2) (i) Six marker, Don't ask.
(ii) m  M = 5log25/10 = 0.0097
3)(i) T = 16111K
(ii) The star was actually a white dwarf but I messed up and put main sequence
(iii) H balmer lines because the star's atmosphere was hot enough to have hydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmer series in the absorption spectra of stars with this surface temperature.
4)(i) A black hole is an object whose escape velocity is greater than c
(ii) Schwarzschild Radius, can't remember.
(iii) 4.96x10 17 s 
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 3727
 19062015 20:34
(Original post by TheRAG)
Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11
(ii) Show that theta = 3.12x10 6 radians
(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)
(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit
2) (i) Six marker, Don't ask.
(ii) m  M = 5log25/10 = 0.0097
3)(i) T = 16111K
(ii) The star was actually a white dwarf but I messed up and put main sequence
(iii) H balmer lines because the star's atmosphere was hot enough to have hydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmer series in the absorption spectra of stars with this surface temperature.
4)(i) A black hole is an object whose escape velocity is greater than c
(ii) Schwarzschild Radius, can't remember.
(iii) 4.96x10 17 s 
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 3728
 19062015 20:37
(Original post by adamjhodg)
I cannot remember the question well enough to know I read it correctly but can you explain your working out for 1i) please? 
Jamesbe1993
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 19062015 20:38
(Original post by TheRAG)
Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit2) (i) Six marker, Don't ask.(ii) m  M = 5log25/10 = 0.00973)(i) T = 16111K(ii) The star was actually a white dwarf but I messed up and put main sequence(iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in its n = 2 state and the deep prominent absorption lines in theBalmer series in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s
~Apparent magnitude graph shows the variation in apparent magnitude of the eclipsing binaries.
~The deeper troughs show that the less luminous star is eclipsing the more luminous star and visa versa for the shallower troughs.
~The maximum height of the graph occurs when the two star are equidistant from the earth ergo one quarter further in there orbit from an eclipse.
~Period can be calculated as the time between two consecutive troughs of equal depth. Therefore 51=4 days.
~Wavelength analysis graph shows the variation in wavelength of the Balmer line centered on ~628 nm.
~Change in wavelength is a consequence of the Doppler effect when the star is moving towards or away from the earth in its orbit.
~Red shift occurs when the star is moving away from us therefore an increase in the wavelength of the light and visa versa for blue shift.
~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding that answer back into z=v/c.Last edited by Jamesbe1993; 19062015 at 20:40. 
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 3730
 19062015 20:41
(Original post by TheRAG)
I used Pythagoras and square rooted 2.75^2 and added 1 (these were the valued in Au) then multiplied by 1.5x10 11 to convert to meters 
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 3731
 19062015 20:42
1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 51=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m  M = 5log25/10 = 0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s

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 3732
 19062015 20:44
What are the sig fig for astro I did 2 sig fig for the majority :/ I'm guessing I've lost a lot of marks if it's meant to be 3 sig figs

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 3733
 19062015 20:44
(Original post by adamjhodg)
Just realised i used the parsec>metre coversion for it...damn!
It did state that the distances were in AU.
But if you got 2.92 AU, you should get 1 mark. 
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 19062015 20:48
(Original post by TheRAG)
1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 51=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m  M = 5log25/10 = 0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s 
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 3735
 19062015 20:49
(Original post by Jamesbe1993)
The mass given on the Schwartzchild radius question was 1x10^10, therefore a radius of ~3x10^13 m is to be calculated.
Were we supposed to calculate v fir the six marker? 
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 3736
 19062015 20:52
Once again, thanks a lot for your help increating this
1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 51=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m  M = 5log25/10 = 0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s 
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 19062015 20:52
(Original post by TheRAG)
Thanks I remember this
Were we supposed to calculate v fir the six marker?Last edited by Jamesbe1993; 19062015 at 20:56. 
Reverse Swing
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 19062015 20:57
(Original post by TheRAG)
Once again, thanks a lot for your help increating this
1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 51=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m  M = 5log25/10 = 0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s 
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 3739
 19062015 21:01
(Original post by TheRAG)
Once again, thanks a lot for your help increating this
1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 51=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m  M = 5log25/10 = 0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s 
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 3740
 19062015 21:06
(Original post by Reverse Swing)
I did the same thing but that's wrong. How many marks out of 3 do you guys reckon we would get for it?
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