# Acceleration of Proton (Kinetic Energy & relativity)

1. Jan 9, 2012

### HarleyM

1. The problem statement, all variables and given/known data

Calculate the kinetic energy required to accelerate a proton from a rest position to 0.9999c. The mass of the proton is 1.67x10-27

Find the ratio of kinetic energy to the energy of a proton at rest

2. Relevant equations

Erest = mc2
Ek = mc2/√(1-v2/c2)

3. The attempt at a solution

Ok So calculating the rest energy is easy

E= (1.67x10-27)(3x108)2
E= 1.503x10-10

Ekinetic= mc2/√(1-v2/c2)
=((1.67x10-27)(3x108)2)/√(1-0.9999c2/c2)
=1.503x10-10/√(1-0.9998)
= 1.503x10-10/ 0.0141418
= 1x10 -8 J

This doesn't seem like a lot of energy to accelerate something to almost light speed I feel like I am missing something...

even when using Etotal =Erest+EK I get 1x10-8 J

can someone point out my mistake?

Ratio of kinetic energy to rest energy is

1.503x10-10 / 1x10-8

Thanks! Happy monday !

2. Jan 9, 2012

### JaWiB

I think you got that backwards

3. Jan 9, 2012

### HarleyM

98.5 % of the energy is kinetic energy then?

4. Jan 10, 2012

### cupid.callin

Your kinetic energy requires will be E when moving - E when at rest !!

So it will be 10-8 - 1.503*10-10 = 9.85 * 10-9

5. Jan 10, 2012

### ehild

v=0.9999c. 1-v^2/c^2 =1-0.9999^2. It is better to expand it as (1-0.9999)(1+0.9999) = 1.9999 E-4.

$$mc^2=\frac{100 m_0c^2}{\sqrt{1.9999}}$$

KE=mc2-m0c2, about 70 times the rest energy.

ehild

Last edited: Jan 10, 2012
6. Jan 12, 2012

### HarleyM

I can get an answer of about 66 times the rest energy, can you explain this formula a little more?

What is M0 and by subtracting the kinetic from rest energy it gives you the ratio?

7. Jan 12, 2012

### ehild

Usually the rest mass is denoted by m0 and the mass of the moving particle is m.

The kinetic energy is the difference between the energy of the moving particle and energy of the particle in rest.

$$KE=mc^2 - m{_0} c^2=\frac{m{_0} c^2}{\sqrt{1-v^2/c^2}}-m{_0} c^2=m{_0} c^2 (\frac{1}{\sqrt{1-v^2/c^2}}-1)$$

The ratio of the kinetic energy to the energy in rest is

$$\frac{KE}{m{_0} c^2} =\frac{1}{\sqrt{1-v^2/c^2}}-1$$

You might have got different result from mine because of the rounding errors.

ehild

Last edited: Jan 12, 2012