The Student Room Group
In part (a) you just go by the standard definition of the potential: the work done in moving that body from infinity to reach a point in the potential. Clearly the work is negative (the body isn't doing work, it's having work done on it, if that makes sense.) Anyhow, negativity falls from that. Try to put it in your own words/etc.

In part (b) you have Ep(r)E_p(r) and note that this is for a single kilogram. Coincidentally, the potential is also defined as being EpE_p per unit mass. Thus they're equivalent. Pick a point (any point, ideally one easily calculable that doesn't require guessing x/y and actually lies on a grid minor) and find MM of the body causing the potential, Mars in this case.

In part (c) you want the escape velocity. Remember how we defined the potential: work done in moving that body from infinity to reach that point. Now, in this case, you want the work done in moving the body from that point to infinity, which is inevitably going to be positive going by (a). Equate that to the kinetic energy of the body (KE->PE to get to infty) and solve through for vescv_\textrm{esc}. Plug your values in and hey-presto.

For (d) picking a few points and using proportionalities/ratios should be enough. Ep1rE_p\propto\frac{1}{r} and so Ep,1Ep,2=r2r1\frac{E_{p,1}}{E_{p,2}}=\frac{r_2}{r_1} and that sort of thing. (they actually expect you to prove that the property Ep(r)×r=constE_p(r) \times r = \textrm{const} apparently, but yeah.)








For the Google Spiders:
Q8.The graph below shows how the gravitational potential energy, Ep, of a 1.0 kg mass varies
with distance, r, from the centre of Mars. The graph is plotted for positions above the
surface of M
(edited 2 years ago)
Reply 2
Original post by Callicious
In part (a) you just go by the standard definition of the potential: the work done in moving that body from infinity to reach a point in the potential. Clearly the work is negative (the body isn't doing work, it's having work done on it, if that makes sense.) Anyhow, negativity falls from that. Try to put it in your own words/etc.

In part (b) you have Ep(r)E_p(r) and note that this is for a single kilogram. Coincidentally, the potential is also defined as being EpE_p per unit mass. Thus they're equivalent. Pick a point (any point, ideally one easily calculable that doesn't require guessing x/y and actually lies on a grid minor) and find MM of the body causing the potential, Mars in this case.

In part (c) you want the escape velocity. Remember how we defined the potential: work done in moving that body from infinity to reach that point. Now, in this case, you want the work done in moving the body from that point to infinity, which is inevitably going to be positive going by (a). Equate that to the kinetic energy of the body (KE->PE to get to infty) and solve through for vescv_\textrm{esc}. Plug your values in and hey-presto.

For (d) picking a few points and using proportionalities/ratios should be enough. Ep1rE_p\propto\frac{1}{r} and so Ep,1Ep,2=r2r1\frac{E_{p,1}}{E_{p,2}}=\frac{r_2}{r_1} and that sort of thing








For the Google Spiders:
Q8.The graph below shows how the gravitational potential energy, Ep, of a 1.0 kg mass varies
with distance, r, from the centre of Mars. The graph is plotted for positions above the
surface of M

Thanks, that makes sense! Although I'm still slightly confused about 8d, could you possibly do a worked example?
Original post by Sid Nandula
Thanks, that makes sense! Although I'm still slightly confused about 8d, could you possibly do a worked example?

The method I proposed to "prove" the proportionality holds is fine, but they're more interested in you actually calculating the potential at two radii r1,r2r_1,r_2 and showing that they equal the points on the graph (sorry- I should have checked the MS to pick up on this! What they want is for you to use the normal formula of V(r)V(r) and calculate it for at least 2 points on the graph, showing it works.

This is just how I would have done it since it seems less reliant on knowing the mass of the body involved.) Anyhow, here's a graphical example of how it works (code snippet attached.)

TSRtest.png


The example case is the Earth: I've plotted the potential over radius as that black line. The blue points I've given are 5 randomly generated points in rr and I've plotted them up for you. They're sorted in ascending order.

You know that V1rV\propto\frac{1}{r} and hence will find that V1V2=r2r1\frac{V_1}{V_2}=\frac{r_2}{r_1}, thusly V1r1V2=r2\frac{V_1r_1}{V_2} = r_2. Using the first randomly generated point on the curve as r1,V1r_1,V_1, and the V2V_2 for each succeeding randomly generated point, I've used the proportionality to try and calculate r2r_2 for each succeeding randomly generated point. I then plot all the "guesses".

These guesses rely only on the initial point I picked on the curve, and the notion that it is indeed a 1r\frac{1}{r} curve. The guesses match the curve, which implies that potential does indeed scale as 1r\frac{1}{r} (I would hope it would since that's what I gave it to graph in the first place! :lol:)

The reason I'd have chosen this method is because typically I'm not confident in my previous answers- this method avoids the requirement of knowing the mass of the body involved. If you messed up getting the mass, this method wouldn't fault you. That being said, if you got the mass wrong it's unlikely you'd be able to pull this off foolproof anyway :tongue:

Spoiler

(edited 2 years ago)
Reply 4
Original post by Callicious
The method I proposed to "prove" the proportionality holds is fine, but they're more interested in you actually calculating the potential at two radii r1,r2r_1,r_2 and showing that they equal the points on the graph (sorry- I should have checked the MS to pick up on this! What they want is for you to use the normal formula of V(r)V(r) and calculate it for at least 2 points on the graph, showing it works.

This is just how I would have done it since it seems less reliant on knowing the mass of the body involved.) Anyhow, here's a graphical example of how it works (code snippet attached.)

TSRtest.png


The example case is the Earth: I've plotted the potential over radius as that black line. The blue points I've given are 5 randomly generated points in rr and I've plotted them up for you. They're sorted in ascending order.

You know that V1rV\propto\frac{1}{r} and hence will find that V1V2=r2r1\frac{V_1}{V_2}=\frac{r_2}{r_1}, thusly V1r1V2=r2\frac{V_1r_1}{V_2} = r_2. Using the first randomly generated point on the curve as r1,V1r_1,V_1, and the V2V_2 for each succeeding randomly generated point, I've used the proportionality to try and calculate r2r_2 for each succeeding randomly generated point. I then plot all the "guesses".

These guesses rely only on the initial point I picked on the curve, and the notion that it is indeed a 1r\frac{1}{r} curve. The guesses match the curve, which implies that potential does indeed scale as 1r\frac{1}{r} (I would hope it would since that's what I gave it to graph in the first place! :lol:)

The reason I'd have chosen this method is because typically I'm not confident in my previous answers- this method avoids the requirement of knowing the mass of the body involved. If you messed up getting the mass, this method wouldn't fault you. That being said, if you got the mass wrong it's unlikely you'd be able to pull this off foolproof anyway :tongue:

Spoiler



Right, I'll need to read your reply again and see exactly how you are doing it as I'm finding it slightly difficult to understand but I'm sure I'll be able to understand it if I just read what you sent thoroughly. I may have some further questions, just a little heads up, but I really appreciate you taking the time to do this! :smile:

Quick Reply

Latest