# Gravitational potential question

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#1
I'm very confused as to how to do the entirety of question 8. Can someone please explain?

Here is the question:

https://www.physicsandmathstutor.com...ial%2520QP.pdf

Here is the mark scheme:

https://www.physicsandmathstutor.com...ial%2520MS.pdf
0
3 days ago
#2
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 and note that this is for a single kilogram. Coincidentally, the potential is also defined as being 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 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 . Plug your values in and hey-presto.

For (d) picking a few points and using proportionalities/ratios should be enough. and so and that sort of thing. (they actually expect you to prove that the property apparently, but yeah.)

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
Last edited by Callicious; 2 days ago
1
#3
(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 and note that this is for a single kilogram. Coincidentally, the potential is also defined as being 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 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 . Plug your values in and hey-presto.

For (d) picking a few points and using proportionalities/ratios should be enough. and so and that sort of thing

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?
0
2 days ago
#4
(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 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 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.)

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 and I've plotted them up for you. They're sorted in ascending order.

You know that and hence will find that , thusly . Using the first randomly generated point on the curve as , and the for each succeeding randomly generated point, I've used the proportionality to try and calculate 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 curve. The guesses match the curve, which implies that potential does indeed scale as (I would hope it would since that's what I gave it to graph in the first place! )

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 Spoiler:
Show
Code:
import matplotlib.pyplot as plt
import numpy as np
import scipy.constants as const
from matplotlib.patches import Patch

font = {'family' : 'serif',
'weight' : 'normal',
'size'  : 12}
import matplotlib as mpl
mpl.rc('font', **font)
# Set up constants
G = const.G
M = 5.972e24
R = 6371e3

# Set up plot to work off
minmax = [R, 10*R]
x = np.linspace(*minmax,1000)
y = -G*M/x

# Pick example points you're going to be working with
random_r = np.sort(np.random.default_rng().integers(*minmax, size=5))
random_V = -G*M/random_r

"""
V = GM/r: ratio V_1/V_2 = r_2/r_1
We should be able to "predict" r_2 based on V_1/V_2 via
r_2 = (V_1/V_2)*r_1
Try this: if it works, it most certainly is a 1/r plot.
"""
# Predict r_2 based on V(r), assuming V(r) proportionate to 1/r. Redo V for this, too.
random_r2 = np.array([random_V*random_r/random_V[d] for d in range(len(random_V))])
random_V2 = -G*M/random_r2

# Set up graphing environment/etc
fig, ax = plt.subplots(nrows=1,ncols=1,squeeze=True)
ax.set(xlabel=r'$r$',
ylabel=r'$V(r)$';)
ax.plot(x,y,color="black",lw=1)
ax.scatter(random_r,random_V,marker="o",s=50,color="blue";)
ax.scatter(random_r2,random_V2, marker="x",s=25, color="yellow";)
legend_elements = [Patch(edgecolor="black",facecolor="blue",label="True";),
Patch(edgecolor="black",facecolor="yellow", label="Guess";)]
plt.legend(loc="lower right",handles=legend_elements)
plt.savefig("TSRtest.png", dpi=300)
plt.show()

Last edited by Callicious; 2 days ago
0
#5
(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 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 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.)

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 and I've plotted them up for you. They're sorted in ascending order.

You know that and hence will find that , thusly . Using the first randomly generated point on the curve as , and the for each succeeding randomly generated point, I've used the proportionality to try and calculate 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 curve. The guesses match the curve, which implies that potential does indeed scale as (I would hope it would since that's what I gave it to graph in the first place! )

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 Spoiler:
Show
Code:
import matplotlib.pyplot as plt
import numpy as np
import scipy.constants as const
from matplotlib.patches import Patch

font = {'family' : 'serif',
'weight' : 'normal',
'size'  : 12}
import matplotlib as mpl
mpl.rc('font', **font)
# Set up constants
G = const.G
M = 5.972e24
R = 6371e3

# Set up plot to work off
minmax = [R, 10*R]
x = np.linspace(*minmax,1000)
y = -G*M/x

# Pick example points you're going to be working with
random_r = np.sort(np.random.default_rng().integers(*minmax, size=5))
random_V = -G*M/random_r

"""
V = GM/r: ratio V_1/V_2 = r_2/r_1
We should be able to "predict" r_2 based on V_1/V_2 via
r_2 = (V_1/V_2)*r_1
Try this: if it works, it most certainly is a 1/r plot.
"""
# Predict r_2 based on V(r), assuming V(r) proportionate to 1/r. Redo V for this, too.
random_r2 = np.array([random_V*random_r/random_V[d] for d in range(len(random_V))])
random_V2 = -G*M/random_r2

# Set up graphing environment/etc
fig, ax = plt.subplots(nrows=1,ncols=1,squeeze=True)
ax.set(xlabel=r'$r$',
ylabel=r'$V(r)$';)
ax.plot(x,y,color="black",lw=1)
ax.scatter(random_r,random_V,marker="o",s=50,color="blue";)
ax.scatter(random_r2,random_V2, marker="x",s=25, color="yellow";)
legend_elements = [Patch(edgecolor="black",facecolor="blue",label="True";),
Patch(edgecolor="black",facecolor="yellow", label="Guess";)]
plt.legend(loc="lower right",handles=legend_elements)
plt.savefig("TSRtest.png", dpi=300)
plt.show()
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! 0
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