# Why is the gravitational potential against distance graph a curve?Watch

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#1
The equation is V =-GM/r so wouldn't this mean that the relationship is V ∝ 1/r which is a straight line not a curve.

Also, I need an explanation for this question:

Which one of the following graphs correctly shows the relationship between the gravitational force, F, between two masses and their separation r.

The answer is D but I don't understand why C can't be the answer as well. All help is appreciated thanks!
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2 years ago
#2
F=GMm/(r^2)

Rewriting this a little u get

F=GMm*(1/r^2)

Comparing this to y=mx c u get he gradient as GMm, the intercept is the origin and the x axis must be 1/r^2.

For a graph as in c, consider

R^2=0.5, F=2GMm
R^2=1, F=GMm
R^2=2, F=GMm/2

As r^2 approaches zero F approaches infinity therefore the graph is not linear. In other words for a double in r^2, F halves.

Try plotting the above values on a graph and u should see it is non linear.

For ur first question u will get a straight line graph, try rearranging for urself as i have done above and comparing to equation of straight line.
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#3
(Original post by Shaanv)
F=GMm/(r^2)

Rewriting this a little u get

F=GMm*(1/r^2)

Comparing this to y=mx c u get he gradient as GMm, the intercept is the origin and the x axis must be 1/r^2.

For a graph as in c, consider

R^2=0.5, F=2GMm
R^2=1, F=GMm
R^2=2, F=GMm/2

As r^2 approaches zero F approaches infinity therefore the graph is not linear. In other words for a double in r^2, F halves.

Try plotting the above values on a graph and u should see it is non linear.

For ur first question u will get a straight line graph, try rearranging for urself as i have done above and comparing to equation of straight line.
I know that it's a straight line I just don't understand why it has to be one but not the other. The D graph shows the line touching 0.

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2 years ago
#4
(Original post by AspiringUnderdog)
The equation is V =-GM/r so wouldn't this mean that the relationship is V ∝ 1/r which is a straight line not a curve.

Also, I need an explanation for this question:

Which one of the following graphs correctly shows the relationship between the gravitational force, F, between two masses and their separation r.

The answer is D but I don't understand why C can't be the answer as well. All help is appreciated thanks!
Consider a graph of y=1/x let v=y and x=r and then -G and M are transformations

you wrote v a 1/r so v=k/r k is -GM
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#5
(Original post by bruh2132)
Consider a graph of y=1/x let v=y and x=r and then -G and M are transformations

you wrote v a 1/r so v=k/r k is -GM
I honestly don't understand what you mean. Sorry
0
2 years ago
#6
(Original post by AspiringUnderdog)
I honestly don't understand what you mean. Sorry
Do you do A level maths/ do you know what the graph of 1/x is?

Also At GCSE Maths you might have learned that if something is proportional to something else you can write an = if you add a k

i.e x ppt y so x=ky
or x ppt 1/y so x=k/y
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#7
(Original post by bruh2132)
Do you do A level maths/ do you know what the graph of 1/x is?

Also At GCSE Maths you might have learned that if something is proportional to something else you can write an = if you add a k

i.e x ppt y so x=ky
or x ppt 1/y so x=k/y
Okay actually I do get what you mean. Having a graph of V against r and plotting V=1/r would be a curve. Holy crap lol I can't believe that I missed this.

So for the graph question. F against r^2 should be a curve so it is incorrect.
Thanks!
0
2 years ago
#8
(Original post by AspiringUnderdog)
Okay actually I do get what you mean. Having a graph of V against r and plotting V=1/r would be a curve. Holy crap lol I can't believe that I missed this.

So for the graph question. F against r^2 should be a curve so it is incorrect.
Thanks!
f against r^2 would be some form of curve yes but we have
f ppt 1/r^2 so if you let 1/r^2 be let's say y
you have f ppt y which would be a straight line if you plot f against y which is f against 1/r^2 if you get what I'm saying

basically f ppt 1/r^2 so there should be a straight line through the origin.
Normally you should consider y=mx+c to see if the graph passes through the origin but here c=0.
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2 years ago
#9
Because you cannot tell if the graph would go through the origin if it was swapped to 1/r^2
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