# Energy to accelerate an 1kg object to 0.9965c?

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1. Mar 28, 2017

### Robert23

1. The problem statement, all variables and given/known data
How much energy does it take to accelerate an object of mass 20,000 kg from rest to 0.9965c?
m = 20,000 kg
v0 = 0 m/s
vf = 0.9965c

2. Relevant equations
$E^{2} = \sqrt{(pc)^{2}+(mc^{2})^2} \\ E_{KE} = (\gamma - 1)mc^{2} \\ \gamma = \sqrt{\frac{1}{1-(v/c)^{2}}} \\ p = \gamma mv$
3. The attempt at a solution
I was unsure if I should use an initial energy and a final energy then take their difference or simply evaluate the relativistic kinetic energy. In doing so I would use the first equation listed for an initial case of v0 = 0 so it's just the objects rest mass energy. The final energy would have vf instead. Then the difference I found was 1.968*10^(22) Joules. However I thought it would also make sense that I can just say the energy to accelerate it from rest is simply the relativistic kinetic energy. Now I evaluated this and found an energy of 2.15*10^(22) Joules. What I don't understand is why these methods do not both work. Is it simply sig-figs or am I missing something? Please let me know if there are any mistakes and which equation is appropriate and why. Thanks!

2. Mar 29, 2017

### ehild

Your first equation should be be
$E = \sqrt{(pc)^{2}+(mc^{2})^2}$ and it is identical to $E=\gamma m c^2$.
It is simpler and equally good to use the equation for the KE $E_{KE} = (\gamma - 1)mc^{2}$ to get the work. To calculate the energy with the first equation is cumbersome and leads to rounding errors.
Show your work in detail. What did you get for γ?

3. Mar 29, 2017

### Robert23

Thanks for the reply. So for the first case I start by solving for gamma:
$\gamma = \sqrt{\frac{1}{1-(.9965)^2}} = 11.96$
Now I can find the kinetic energy:
$KE = (11.96-1)(20,000kg)(c)^2 , c = 3*10^{8} m/s$
KE = 2.138*10^{22} J

Now for the other method using total energies, I first find the momentum:
$p = 11.96(20,000 kg)(0.9965c) = 7.146*10^{13} kg*m/s$
So now I can solve for $E_{final}$:
$E_{final} = \sqrt{(pc)^{2}+(mc^{2})^2}$
$E_{final} = \sqrt{(7.146*10^{13} kg*m/s * 3*10^{8} m/s)^2 + (20,000kg*(3*10^8 m/s)^2)^2}$
$E_{final} = 2.151*10^{22} J$
and to find $E_{initial}$:
$E_{initial} = (20,000kg)(3*10^8 m/s)^2$
$E_{initial} = 1.80*10^{21} J$
therefore $\Delta E = 1.971*10^{22} J$

So are these just rounding errors? They're 0.167 apart which feels rather large to me.

4. Mar 29, 2017

### ehild

It must be rounding error as the two formulas for the total energy are identical in principle. The speed is given for four significant digits. Use at least 5 digits in the calculations.

5. Mar 29, 2017

### OmCheeto

Nope. I just plugged the numbers into a spreadsheet, and came up with a similar error.

Funny thing is, if you replace the speed of light, from 3e8 m/s with 1, and take the difference of the two answers, you come up with a "funny" number: 20,000.

hmmmm......

I don't have relativistic equations memorized, so I will have to figure out why [√((pc)^2+(mc^2)^2)] - [(γ-1)*mc^2] = m

6. Mar 29, 2017

### PeroK

$E^2 = p^2c^2+ m^2c^4 = \gamma^2 m^2v^2c^2+m^2c^4 = \gamma^2m^2c^2(v^2+\frac{c^2}{\gamma^2}) = \gamma^2m^2c^2(c^2)$

7. Mar 29, 2017

### PeroK

The first term is total energy, the second is kinetic energy. They differ by $mc^2$.

8. Mar 30, 2017

### ehild

You simply forgot to subtract 1 from gamma.
$KE = (11.96-1)(20,000kg)(c)^2 =1.9728\cdot 10^{22}$

9. Apr 2, 2017

### Robert23

AHHHH thank you!