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Acceleration of Proton (Kinetic Energy & relativity)
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ehild
#
5
Jan10-12, 04:19 AM
HW Helper
Thanks
P: 10,760
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.
[tex]mc^2=\frac{100 m_0c^2}{\sqrt{1.9999}}[/tex]
KE=mc
^{2}
-m
_{0}
c
^{2}
, about 70 times the rest energy.
ehild