Finding speed from acceleration through a potential difference

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Homework Help Overview

The discussion revolves around finding the speed of a proton in terms of the speed of light (c) based on its acceleration through a potential difference of 77 MV. The original poster attempts to relate potential energy and relativistic concepts to derive the speed but encounters confusion regarding the gamma factor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of potential energy and relativistic kinetic energy to find the speed of a proton. Questions arise about the interpretation of gamma and its implications when calculated to be less than one. There is also a shift in focus to a separate question regarding the Schwarzschild radius and the significance of mass order in calculations.

Discussion Status

Some participants have provided guidance on using the correct expressions for relativistic kinetic energy and emphasized the importance of unit consistency in calculations. The original poster acknowledges the feedback and indicates a willingness to revise their approach. The discussion has shifted to another topic, with participants exploring the Schwarzschild radius and addressing unit conversion issues.

Contextual Notes

There is mention of homework constraints, such as the requirement for specific units and significant figures in the context of a webassign quiz. The original poster expresses concern about entering answers correctly based on these constraints.

metalmagik
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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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One should use the expression for relativistic kinetic energy;

[tex]E_{k}=m_{0}c^2(\gamma - 1)[/tex]
 
but I need to find V, velocity.
 
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
 
thanks very much, I got the question right. I have one more question if you are still around.
 
No problem, I'm still here.
 
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.
 
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.
 
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
 

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