Finding speed from acceleration through a potential difference

In summary, the conversation revolved around finding the speed of a proton in terms of the speed of light when given an acceleration through a potential difference of 77 MV. The solution involved using the expressions for relativistic kinetic energy and finding gamma, which was found to be less than one. The conversation then shifted to finding the Schwarzschild radius for a galaxy with a mass on the order of 10^41 kg, which required being careful with units and significant figures. The correct answer was eventually found and the person was able to achieve a 100% score on their quiz.
  • #1
metalmagik
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0
I need help finding the speed of a proton in terms of c (speed of light) when all I have is an acceleration through a potential difference of 77 MV.

I started by saying this was 77,000,000 Volts, and then using PE = qV to solve for energy. I used E = mc^2 to find mass and used this and rest mass of a proton to find gamma by using m = mo(gamma). I found gamma to be less than one. How is this possible? What did I do wrong? Any help is greatly appreciated.
 
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  • #2
One should use the expression for relativistic kinetic energy;

[tex]E_{k}=m_{0}c^2(\gamma - 1)[/tex]
 
  • #3
but I need to find V, velocity.
 
  • #4
Oh wait, I see what you're saying, I will try this and post again if I have not done it correctly again. Thank you
 
  • #5
thanks very much, I got the question right. I have one more question if you are still around.
 
  • #6
No problem, I'm still here.
 
  • #7
Great, I was going to post a new topic but I didn't want to flood the board too much.

What is the Schwarzschild radius for a typical galaxy (with mass on the order of 10^41 kg)?

I do not understand the "with mass on the order of 10^41 kg"

I tried plugging (1e41) with the Schwarzschild radius formula but I got the wrong answer when I entered it into the test. I have one more response left so I need to get it right, thanks so much for the help so far, I'd be grateful if you could work with me on this one.
 
  • #8
On the order of means exactly what you did. Off the order usually means powers of ten. If this is a webassign homework you probably know already, but you need to be very careful with the way you enter your answers (significant figures, units metres of kilometres etc). I've just done a quick calc myself and arrived at a radius of about 1.48x1014m.
 
  • #9
Ah! that must be why! I got the same exact answer, but they want it in km. I did not realize the Schwarzschild radius formula is worked to give answers in meters. So, since my webassign quiz wants it in km, it should be, 1.48x10^11 km, correct?
 
  • #10
metalmagik said:
Ah! that must be why! I got the same exact answer, but they want it in km. I did not realize the Schwarzschild radius formula is worked to give answers in meters. So, since my webassign quiz wants it in km, it should be, 1.48x10^11 km, correct?
If your unsure check your units, assuming you entered everything into the equation in SI units;

[tex]R = \frac{2GM}{c^{2}} =\frac{(m^{3}kg^{-1}s^{-2})(kg)}{(ms^{-1})^2} = m[/tex]

Assuming webassign wants it to 3sf then yes, your answer should be correct.
 
  • #11
got it right! 100% score wahoo

thanks a lot Hootenanny, thanks verryyy much for your time.
 
  • #12
metalmagik said:
got it right! 100% score wahoo
Well done mate. :approve:
metalmagik said:
thanks a lot Hootenanny, thanks verryyy much for your time.
My pleasure
 

FAQ: Finding speed from acceleration through a potential difference

What is the formula for finding speed from acceleration through a potential difference?

The formula for finding speed from acceleration through a potential difference is: Speed = √2ad, where a is the acceleration and d is the potential difference.

How is acceleration related to potential difference?

Acceleration is directly proportional to potential difference. This means that as the potential difference increases, the acceleration also increases.

Can speed be calculated without knowing the potential difference?

No, speed cannot be calculated without knowing the potential difference. It is a crucial factor in the formula for finding speed from acceleration through a potential difference.

How does increasing the potential difference affect the speed?

Increasing the potential difference will result in an increase in speed. This is because a higher potential difference means a greater force acting on the object, causing it to accelerate faster and reach a higher speed.

Is there a limit to how fast an object can go using this method?

Yes, there is a limit to how fast an object can go using this method. This is because as the object approaches the speed of light, its mass increases, making it more difficult to accelerate further.

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