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anonymous
 one year ago
Determine the ratio of the relativistic to the non relativistic kinetic energy when V=1.5x10^3 m/s and when V = 0.97 m/s.
I have tried it three times and for some reason my answer is always zero
anonymous
 one year ago
Determine the ratio of the relativistic to the non relativistic kinetic energy when V=1.5x10^3 m/s and when V = 0.97 m/s. I have tried it three times and for some reason my answer is always zero

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[KE _{rel}=mc ^{2}(\gamma1)\] \[KE _{nonrel}=\frac{ 1 }{ 2 }mV ^{2}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Looks like fun, can you show your work please?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\gamma = \frac{ 1 }{ \sqrt{1\left( \frac{ v }{ c } \right)}^2 }\] just to note the Lorentz factor

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0It's probably best if you do the calculations separately

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[KE_{rel} = mc^2\left( \frac{ 1 }{ \sqrt{1\left( \frac{ v }{ c } \right)^2} }1 \right)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0You should find what fraction of the speed of light the given velocities are and then you can find the ratio.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So \[1.5 \times 10^{3} m/s = \frac{ 1 }{ 450,000 } c \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\gamma = \frac{ 1 }{ \sqrt{1\left( \frac{ (1/450,000)c)^2 }{ c^2 } \right)} }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0The reason you're getting 0 is, when you subtract by 1 using a calculator it's approximating to 1 as the numbers are so small, we get something insane such as 0.9999999...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for small velocities (v<<c), we can write this (using Taylor expansion): \[\Large E = \gamma m{c^2} \cong m{c^2}\left( {1 + \frac{{{v^2}}}{{2{c^2}}}} \right) = m{c^2} + m\frac{{{v^2}}}{2}\] so we can write: \[\Large K{E_{rel}} = \gamma m{c^2}  m{c^2} \cong m\frac{{{v^2}}}{2} = K{E_{not\;rel}}\] so, in the case v << c, your ratio is equal to 1
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