Finding Escape Velocity of mass (m) with electrostatic and gravitational influences

  • #1
PhysicsEnjoyer31415
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Sorry moderators i am still learning LaTex so i had no other choice

So i tried calculating escape velocity for a charged body of mass (m) and charge (+q) when a charged body (-q) is present at a distance (d) and kept stationary. I equated the Gravitational potential and electrostatic potential energies for initial and final and i got this expression(without any approximation for x<<d or d tends to infinity) . I would like to know if what i have derived is correct or have i done something wrong
20240522_230623.jpg
 
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  • #2
How do you define escape velocity? It looks like you messed up the energy conservation equation. On the left you have the initial configuration when the mass starts at d from the charge with initial speed ##v_e##. On the right you have an expression saying that the mass is closer to the charge and has zero speed. How can that be if there is an attractive force between the mass and the the charge? The speed should increase as the two get closer together.

Also, both electric and gravitational potential energies should be written as if all the mass and all the charge of the charged body were concentrated at its center.

Also, why are there two gravitational terms, ##-\dfrac{GMm}{d}## and ##-\dfrac{GMm}{R}## on the left hand side?

Also, you will be well-advised to ignore the gravitational energy. ##G## is 21 orders of magnitude lower than ##k##. If you ignore gravity, your expression gives an imaginary number for the escape velocity.
 
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  • #3
PhysicsEnjoyer31415 said:
i am still learning LaTex so i had no other choice
No other choice? Really? To me it sounds like "If you are visually impaired, I don't want your help or participation."

Hope that works out for you.
 
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  • #4
Gotta love the combination square root and answer box, though. I think that's the first time I've seen that. :smile:

1716412393600.png
 
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  • #5
kuruman said:
How do you define escape velocity? It looks like you messed up the energy conservation equation. On the left you have the initial configuration when the mass starts at d from the charge with initial speed ##v_e##. On the right you have an expression saying that the mass is closer to the charge and has zero speed. How can that be if there is an attractive force between the mass and the the charge? The speed should increase as the two get closer together.

Also, both electric and gravitational potential energies should be written as if all the mass and all the charge of the charged body were concentrated at its center.

Also, why are there two gravitational terms, ##-\dfrac{GMm}{d}## and ##-\dfrac{GMm}{R}## on the left hand side?

Also, you will be well-advised to ignore the gravitational energy. ##G## is 21 orders of magnitude lower than ##k##. If you ignore gravity, your expression gives an imaginary number for the escape velocity.
The two gravitational terms were for potential of the mass towards planet and potential of mass towards charge.But yeah i did not see the imaginary number thing .Ill try to improve the formula again by today night
 
  • #6
berkeman said:
Gotta love the combination square root and answer box, though. I think that's the first time I've seen that. :smile:

View attachment 345752
Oh that was by accident , i did not even notice that till now sorry haha
 
  • #7
Vanadium 50 said:
No other choice? Really? To me it sounds like "If you are visually impaired, I don't want your help or participation."

Hope that works out for you.
i thought my handwriting would not be clear enough to understand , sorry
 
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  • #8
berkeman said:
Gotta love the combination square root and answer box, though. I think that's the first time I've seen that. :smile:

1716412393600-png.png
Yeah, try to do that with your LaTex!
 
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  • #9
PhysicsEnjoyer31415 said:
The two gravitational terms were for potential of the mass towards planet and potential of mass towards charge.But yeah i did not see the imaginary number thing .Ill try to improve the formula again by today night
Planet? What planet? At that scale gravity will dominate any realistic electrostatic force.

You would be better taking a fixed central charge of ##Q## and a charge ##q## of mass ##m##. Where the charges have opposite signs.
 
  • #10
PhysicsEnjoyer31415 said:
I equated the Gravitational potential and electrostatic potential energies
Why not equate the (negated) total potential energy (using infinity as reference) to kinetic energy?
 
  • #11
A.T. said:
Why not equate the (negated) total potential energy (using infinity as reference) to kinetic energy?
Isnt that what i did , the gravitational potential energy of the mass towards both planet and charge?
 
  • #12
PeroK said:
Planet? What planet? At that scale gravity will dominate any realistic electrostatic force.

You would be better taking a fixed central charge of ##Q## and a charge ##q## of mass ##m##. Where the charges have opposite signs.
But the question is to find escape velocity in this very scenario , i might try this in another post if i get the time .Thank you for the Idea kind stranger
 
  • #13
PhysicsEnjoyer31415 said:
But the question is to find escape velocity in this very scenario , i might try this in another post if i get the time .Thank you for the Idea kind stranger
If this is homework it should be in the homework section. In any case, you need to post a full problem statement.
 
  • #14
PeroK said:
If this is homework it should be in the homework section. In any case, you need to post a full problem statement.
Not homework, i create personal problems to keep my physics sharp and derive some stuff for fun , its one of them
 
  • #15
PhysicsEnjoyer31415 said:
Not homework, i create personal problems to keep my physics sharp and derive some stuff for fun , its one of them
That still counts as homework. All the more reason to post a full problem statement.
 
  • #16
PeroK said:
That still counts as homework. All the more reason to post a full problem statement.
Well if you want it really well defined , the problem statement could be "Find escape velocity if charge -q is stationary" i guess?.But how does a derivation count as homework though?
 
  • #17
PhysicsEnjoyer31415 said:
Well if you want it really well defined
Of course we do! If it's not well defined, how do you expect us to understand it?

What have you told us so far?
  • It's too much trouble for you to use LaTex. (And if you are visually impaired, <blank> you)
  • Its our responsibility to figure out your writing, not yours to post clearly
  • Despite PF Rules, you don't want to put homework-type questions in the homework forums.
  • Posting a clear, well-defined question is too much trouble.
And you want our help. Does this sound like a winning strategy to get it?
 
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  • #18
Vanadium 50 said:
Of course we do! If it's not well defined, how do you expect us to understand it?

What have you told us so far?
  • It's too much trouble for you to use LaTex. (And if you are visually impaired, <blank> you)
  • Its our responsibility to figure out your writing, not yours to post clearly
  • Despite PF Rules, you don't want to put homework-type questions in the homework forums.
  • Posting a clear, well-defined question is too much trouble.
And you want our help. Does this sound like a winning strategy to get it?
I am sorry i will post these type on the homework from next time
 

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