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

I did this? but I kept getting 9. something ?

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

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

$\therefore \frac{g_s}{g_p}=\frac{g_s}{8}=\frac{M_s}{M_p}\left(\frac{r_p}{r_s}\right)^2$.

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

$\frac{\rho_p}{\rho_s}=1=\frac{M_p}{M_s}\left(\frac{r_s}{r_p}\right)^3=100^3\frac{M_p}{M_s}$, 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.

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?

13 was fine haha it was 14 i was after

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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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

$v=\sqrt{\frac{GM}{r}}\rightarrow v^2\propto \frac{1}{r}$.

Ohhhhhh! So youre talking about potential! Sorry the lower case v is normally velocity I'm with you now!

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!

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).

And last question for anyone out there who understands
http://filestore.aqa.org.uk/subjects/AQA-PHYA4-2-QP-JUN12.PDF
question 4biii
didnt really understand the mark scheme on that one

The formula for gravitational potential is:

$[br]V = (-) \dfrac{GM}{r}[br]$

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

And last question for anyone out there who understands
http://filestore.aqa.org.uk/subjects/AQA-PHYA4-2-QP-JUN12.PDF
question 4biii
didnt really understand the mark scheme on that one