Finding gravitational potential and intensity in point A

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  • #1
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Homework Statement


Here is the pic: http://i.imgur.com/olnuDjL.jpg


Homework Equations





The Attempt at a Solution



So intensity was pretty easy, it came up to be [tex]y=\frac{GM}{a^{2}}(1+\frac{1}{2\sqrt{2}},1+\frac{1}{2\sqrt{2}})[/tex] Check on if it's correct would be nice aswell. Now for potential I know that
[tex]y=-\nabla V[/tex]
so that means
[tex]y=-(\frac{dV}{dx},\frac{dV}{dy})[/tex]

and I have troubles solving that, mainly because I don't know how to handle dy and dx.

I assume

[tex]V=\int \frac{GM}{a^2}(1+\frac{1}{2\sqrt{2}}) da = -\frac {GM}{a} (1+\frac{1}{2\sqrt{2}})[/tex]

but I don't know why
 
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Answers and Replies

  • #2
tms
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So intensity was pretty easy, it came up to be [tex]y=\frac{GM}{a^{2}}(1+\frac{1}{2\sqrt{2}},1+\frac{1}{2\sqrt{2}})[/tex] Check on if it's correct would be nice aswell.
That's not what I get (although I could have made a mistake).

Now for potential I know that
[tex]y=-\nabla V[/tex]
so that means
[tex]y=-(\frac{dV}{dx},\frac{dV}{dy})[/tex]

and I have troubles solving that, mainly because I don't know how to handle dy and dx.
There is a simple, direct expression for the potential.
 
  • #3
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That's not what I get (although I could have made a mistake).


There is a simple, direct expression for the potential.

Well what did you come up with for intensity then?
 
  • #4
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I also got something different. The way this forum works is you show us your solution (not just the answers), and we point out possible mistakes. What's the formula for the potential of a single point mass? You will need that to find the potential of that system. Just keep in mind that the potential is additive - that is you can add the potential due to each individual mass to find the total potential of the whole system.
 
  • #5
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If you say you got something different then atleast write your different answer, writing everything takes a lot of time in latex

[tex]y=\frac{GM}{a^{2}}*(1,0)+\frac{GM}{a^{2}}*(0,1)+\frac{GM}{2a^{2}}*( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})[/tex]

What about the potential? What is that "simple" formula?
 
  • #6
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Actually, I think at first I misunderstood your picture. Now I think I understand it better and agree with your answer except for a minus sign due to the attractive force. Don't you know the formula for gravitational potential energy?
 
  • #7
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Do you mean minus sign in y? Why? Vectors specify the direction and they are clearly all positive.

I only know whe formula with gradient, and that's what I should use here, can you just tell me how to properly calculate it?
 
  • #8
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I mean a minus sign in both y and x directions. The forces are attractive so the forces point to the masses that produce them, not away from them. The potential due to a mass m at point a distance r away from the mass is given by ø = -Gm/r
 
  • #9
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Uh, but I have to find potential of this ... system? Not for each of the masses I think, that's why I thought this formula isn't good to use here.
 
  • #10
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Uh, but I have to find potential of this ... system? Not for each of the masses I think, that's why I thought this formula isn't good to use here.

Potential is an additive quantity. Find the potentials due to each individual mass and add them up.
 
  • #11
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Ok, can you explain to me (preferably with some example) when should I use the formula with gradient vs this simpler one? And if here I could use the gradient one, how could I make it work?
 
  • #12
tms
644
17
If you say you got something different then atleast write your different answer, writing everything takes a lot of time in latex
LaTeX is the same for everyone, save that some have more experience with it. At any rate, the rules of Physics Forums prohibit just handing out answers; the one asking for help must show his or her work.

[tex]y=\frac{GM}{a^{2}}*(1,0)+\frac{GM}{a^{2}}*(0,1)+\frac{GM}{2a^{2}}*( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})[/tex]
Yes, that is correct, except for the sign. I made an unbelievably stupid algebra error. Sorry. It would also be helpful if you used standard symbols for things; ##y## is usually a part of a position, not the electric field.
 

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