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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

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    (Original post by Leonacatherine)
    I did this? but I kept getting 9. something ?
    

v = \sqrt{\dfrac{6.67 \times 10^{-11} \times 1.99 \times 10^{30}}{1.58 \times 10^{11}}}


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    13) Let subscripts p/s denote a property of the planet/star respectively.

    g\propto \frac{M}{r^2}

    \therefore \frac{g_s}{g_p}=\frac{g_s}{8}=\f  rac{M_s}{M_p}(\frac{r_p}{r_s})^2.

    Moreover, \rho_s=\rho_p and \rho \propto\frac{M}{r^2}, so:

    \frac{\rho_p}{\rho_s}=1=\frac{M_  p}{M_s}(\frac{r_s}{r_p})^3=100^3  \frac{M_p}{M_s}, since we are given that d_s=100d_p, and d\propto r.

    Solving this last equation we have \frac{M_s}{M_p}=100^3 and plugging into the previous equation we get \frac{g_s}{8}=100^3\cdot \frac{1}{100^2}=100, therefore g_s=800\text{Nkg}^{-1}.

    Edit: Sorry about the latex, I did try it's just that the latex on this site really needs fixing up.

    Edit2: Removed \left and \right
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    (Original post by PotterPhysics)
    13) Let subscripts p/s denote a property of the planet/start respectively.

    g\propto \frac{M}{r^2}

    \therefore \frac{g_s}{g_p}=\frac{g_s}{8}=\f  rac{M_s}{M_p}\left(\frac{r_p}{r_  s}\right)^2.

    Moreover, \rho_s=\rho_p and \rho \propto\frac{M}{r^2}, so:

    \frac{\rho_p}{\rho_s}=1=\frac{M_  p}{M_s}\left(\frac{r_s}{r_p}\rig  ht)^3=100^3\frac{M_p}{M_s}, since we are given that d_s=100d_p, and d\propto r.

    Solving this last equation we have \frac{M_s}{M_p}=100^3 and plugging into the previous equation we get \frac{g_s}{8}=100^3\cdot \frac{1}{100^2}=100, therefore g_s=800\text{Nkg}^{-1}.


    Edit: Sorry about the latex, I did try it's just that the latex on this site really needs fixing up.
    13 was fine haha it was 14 i was after
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    (Original post by Leonacatherine)
    I dont follow your answer
    I understand this bit
    so .

    up to the square root but dont get how youve introduced t and another m?
    It's just the appalling latex on this site, I didn't introduce any t.
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    (Original post by Leonacatherine)
    13 was fine haha it was 14 i was after
    Kinetic energy and speed are inversely proportional to radius (and radius squared) respectively (the closer the body to the surface of a planet, the greater its KE and greater it's speed).

    Potential energy and time period (squared) are proportional to the radius (and radius cubed) respectively (the closer the body to the surface of a planet, the smaller it's PE and shorter its time period).

    The radius of Q is 3 times the radius of P.

    This follows that the only possible correct option is A, using the above reasoning.





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    (Original post by Leonacatherine)
    13 was fine haha it was 14 i was after
    You wrote 13 -- I've just deleted my answer to 14 because CD's answer is better (more instructive).
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    (Original post by CD223)
    Kinetic energy and speed are inversely proportional to radius (and radius squared) respectively (the closer the body to the surface of a planet, the greater its KE and greater it's speed).

    Potential energy and time period (squared) are proportional to the radius (and radius cubed) respectively (the closer the body to the surface of a planet, the smaller it's PE and shorter its time period).

    The radius of Q is 3 times the radius of P.

    This follows that the only possible correct option is A, using the above reasoning.





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    See i thought! But if ke = 1/2 mv^2 and v =2 x pi x radius x frequency then wouldnt ke increase with radius according to that equation?
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    On second thoughts here's an alternative to 14.

    14) Since the masses are the same, E_k\propto v^2 and v^2\propto \frac{1}{r}. Therefore E_k\propto \frac{1}{r}. It follows that \frac{E_{k(P)}}{E_{k(Q)}}=\frac{  r_q}{r_p}=3, \therefore E_{k(P)}=3E_{k(Q)}>E_{k(Q)}. So the answer is A.
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    (Original post by PotterPhysics)
    On second thoughts here's an alternative to 14.

    14) Since the masses are the same, E_k\propto v^2 and v^2\propto \frac{1}{r}. Therefore E_k\propto \frac{1}{r}. It follows that \frac{E_{k(P)}}{E_{k(Q)}}=\frac{  r_q}{r_p}=3, \therefore E_{k(P)}=3E_{k(Q)}>E_{k(Q)}. So the answer is A.
    i dont get how is v squared proportional to 1/r ? which equation is this from
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    Ive only just realised I'm talking to two different people! haha Potterphysics and cd223 are such brainboxes I merged them into one person lol!
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    (Original post by Leonacatherine)
    i dont get how is v squared proportional to 1/r ? which equation is this from
    v=\sqrt{\frac{GM}{r}}\rightarrow v^2\propto \frac{1}{r}.
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    (Original post by PotterPhysics)
    v=\sqrt{\frac{GM}{r}}\rightarrow v^2\propto \frac{1}{r}.
    Ohhhhhh! So youre talking about potential! Sorry the lower case v is normally velocity I'm with you now! And i get why i cant use the 1/2 mv^2 because they arent in a uniform field so I need to use the radial equations right?

    Thanks so much for all your help really appreciate it
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    (Original post by Leonacatherine)
    Ohhhhhh! So youre talking about potential! Sorry the lower case v is normally velocity I'm with you now!
    Absolutely not!

    v is velocity -- the equation is derived from the fact that \text{centripetal force}=\text{gravitational attraction} for an object undergoing circular motion about a mass due to the gravitational pull of the mass. Therefore \frac{mv^2}{r}=\frac{GMm}{r^2} giving v=\sqrt{\frac{GM}{r}}.

    Edit: I should add that E_k\propto v^2 is from E_k=\frac{1}{2}mv^2\propto v^2 (because, as I said, the m's are the same for both satellites).
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    (Original post by Leonacatherine)
    Ive only just realised I'm talking to two different people! haha Potterphysics and cd223 are such brainboxes I merged them into one person lol!
    Lol I'm really not


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    (Original post by PotterPhysics)
    Absolutely not!

    v is velocity -- the equation is derived from the fact that \text{centripetal force}=\text{gravitational attraction} for an object undergoing circular motion about a mass due to the gravitational pull of the mass. Therefore \frac{mv^2}{r}=\frac{GMm}{r^2} giving v=\sqrt{\frac{GM}{r}}.

    Edit: I should add that E_k\propto v^2 is from E_k=\frac{1}{2}mv^2\propto v^2 (because, as I said, the m's are the same for both satellites).
    oh gosh
    right so to put it simply by equating the centripetal force to the gravitational force v^2 is inversely proportional to r so Ek is too. if Ek is inversely proportional to r a larger radius will have a smaller Ek so it has to be A
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    And last question for anyone out there who understands
    http://filestore.aqa.org.uk/subjects...2-QP-JUN12.PDF
    question 4biii
    didnt really understand the mark scheme on that one
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    (Original post by Leonacatherine)
    And last question for anyone out there who understands
    http://filestore.aqa.org.uk/subjects...2-QP-JUN12.PDF
    question 4biii
    didnt really understand the mark scheme on that one
    The formula for gravitational potential is:

    

V = (-) \dfrac{GM}{r}

    All the mark scheme is saying is that:

    • the distance from the Earth to the Sun is much greater than the distance from the Earth to the Moon
    • As such, the change in potential due to the sun is negligible over the distance travelled between the earth and the moon
    • The relative movement of the moon around the earth and the rotation of earth about its axis is so small compared to the distances considered, that the Suns potential can be ignored when working out the change in potential from Earth to point X in order to work out the energy required


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    (Original post by CD223)
    The formula for gravitational potential is:

    

V = (-) \dfrac{GM}{r}

    All the mark scheme is saying is that:

    • the distance from the Earth to the Sun is much greater than the distance from the Earth to the Moon
    • As such, the change in potential due to the sun is negligible over the distance travelled between the earth and the moon
    • The relative movement of the moon around the earth and the rotation of earth about its axis is so small compared to the distances considered, that the Suns potential can be ignored when working out the change in potential from Earth to point X in order to work out the energy required


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    Gotcha! Perfect thank you
    I always get so stressed when I'm marking these but the grade boundaries are so low! ends up fine haha
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    (Original post by Leonacatherine)
    Gotcha! Perfect thank you
    I always get so stressed when I'm marking these but the grade boundaries are so low! ends up fine haha
    Yeah the papers are hard. Physics is hard in general haha!


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    (Original post by Leonacatherine)
    And last question for anyone out there who understands
    http://filestore.aqa.org.uk/subjects...2-QP-JUN12.PDF
    question 4biii
    didnt really understand the mark scheme on that one
    X is the point where GM/r² is zero. You can imagine it to be a see saw. At the middle of the see saw there is no movement, but as I etch forwards towards the moon, I begin to tip towards it just as I pass the centre. This is due to being attracted to it as the gravitational field strength for the moon now exceeds the earth's field. I hope that makes sense


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