Calculate the kinetic energy required to accelerate a proton

[HarleyM]

Homework Statement

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

Homework Equations

E_rest = mc²
E_k = mc²/√(1-v²/c²)

The Attempt at a Solution

Ok So calculating the rest energy is easy

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

E_kinetic= 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 E_total =E_rest+E_K I get 1x10^-8 J

can someone point out my mistake?

Ratio of kinetic energy to rest energy is

1.503x10^-10 / 1x10^-8
= 1.503 % of the energy is kinetic energy ? ( really unsure about this) Thanks! Happy monday !

 

[JaWiB]

Ratio of kinetic energy to rest energy is

1.503x10^-10 / 1x10^-8
= 1.503 % of the energy is kinetic energy ? ( really unsure about this)

I think you got that backwards

 

[HarleyM]

I think you got that backwards

98.5 % of the energy is kinetic energy then?

 

[cupid.callin]

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

E_Moving= 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
​

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

 

[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

 

[HarleyM]

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

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?

 

[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