Relativistic Energy Calculations

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The discussion centers on calculating the total energy of two colliding protons, each with 7 TeV, resulting in a total of 14 TeV. The energy of an Airbus A320, with a corrected mass of 5*10^4 kg and a speed of 600 km/h, is recalculated using the non-relativistic formula, yielding approximately 8.3*10^7 J or 5.2*10^25 eV. Participants clarify that the kinetic energy should be calculated rather than total mass-energy, and the initial mass value was incorrect. The final calculation indicates that approximately 7.4*10^10 protons are needed to match the Airbus's energy. The conversation emphasizes the importance of using proper units and formulas in relativistic energy calculations.
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Homework Statement


Hi everybody. I was tasked with the following exercice: Two identical protons of energy 7TeV collide head on.
Which is the total energy of the system?
How is compared this energy with the one of an Airbus 320 which mass is t50*10^6Kg at a speed of 600Km/h?
How many photons do we need for the total energy was equivalent to the Airbus?

Homework Equations


I am considering that the speed of light is 1 and it has no units.
Relativistic energy: E=Sqrt(m^2+p^2)

The Attempt at a Solution


To the first question: If both photons have the same energy, the total energy of the system must be 14 TeV; exactly twice the energy of the photon.

To the second question: Here, I write the aeroplane's energy as E=Sqrt(m^2+(m*v)^2), qhere v is the speed (all in the SI). So E=Sqrt((50*10^4)^2+((50*10^4)*(166.67))^2) =8.3*10^7J.
On the other hand 1J=1/(1.602*10^19)eV; so E =5.2*10^25eV=5.2*10^11Tev.

To the third question: 5.2/7=0.74; so we need 7.4*10^10 protons to obtain the same energy.


Can someone please tell me if I have calculated the energy of the aeroplane properly?

Thanks for reading.
 
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You mix protons and photons, I guess you always mean protons.
Frank Einstein said:
To the first question: If both photons have the same energy, the total energy of the system must be 14 TeV; exactly twice the energy of the photon.
Correct.

Your Airbus mass changed from 50*106 to 50*104 kg? Both numbers are unrealistic, the mass should be somewhere in the range of 5*104 to 10*104 kg.
Frank Einstein said:
E=Sqrt(m^2+(m*v)^2)
This equation does not work if you use SI units. In the SI, the speed of light is not one.
Frank Einstein said:
Here, I write the aeroplane's energy as E=Sqrt(m^2+(m*v)^2)
That would be the total mass-energy of the airbus, but the question asks for the kinetic energy only (not explicitely, but the question wouldn't make sense otherwise).
 
mfb said:
You mix protons and photons, I guess you always mean protons.
Correct.

Your Airbus mass changed from 50*106 to 50*104 kg? Both numbers are unrealistic, the mass should be somewhere in the range of 5*104 to 10*104 kg.
This equation does not work if you use SI units. In the SI, the speed of light is not one.
That would be the total mass-energy of the airbus, but the question asks for the kinetic energy only (not explicitely, but the question wouldn't make sense otherwise).

Yeah, the mass of the Airbus is 5*10^4kg; it was a mistake.
Then, for the second question should I have written energy as the non relativistic 0.5*m*(v^2)?
 
Frank Einstein said:
Then, for the second question should I have written energy as the non relativistic 0.5*m*(v^2)?
Yes, this is a good approximation as 600 km/h is much much smaller than c.
 
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Orodruin said:
Yes, this is a good approximation as 600 km/h is much much smaller than c.
Allright. I gess that I will do that.

Thak you very much both of you.
 
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