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Chapter 5: Problem 73

(a) At what relative velocity is \(\gamma=2.00 ?\) (b) At what relative velocity is \(\gamma=10.0 ?\)

Short Answer

Expert verified
(a) The relative velocity corresponding to \(\gamma=2.00\) is approximately \(2.60 \times 10^8 m/s\). (b) The relative velocity corresponding to \(\gamma=10.0\) is approximately \(2.99 \times 10^8 m/s\).

Step by step solution

01

Rearrange the Lorentz factor equation for relative velocity v

We have the equation \[\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\] We need to solve it for \(v\), where \(c\) is the speed of light. First, let's square both sides of the equation: \[\gamma^2 = \frac{1}{1-\frac{v^2}{c^2}}\] Now, solve for \(v^2\): \[v^2 = c^2(1-\frac{1}{\gamma^2})\] And finally, solve for \(v\): \[v = c\sqrt{1-\frac{1}{\gamma^2}}\]
02

Calculate the relative velocity for \(\gamma=2.00\)

Now that we have the equation for the relative velocity, let's plug in the Lorentz factor of 2.00: \[v = c\sqrt{1-\frac{1}{(2.00)^2}}\] Calculate the value inside the square root: \[v = c\sqrt{1-\frac{1}{4}}\] \[v = c\sqrt{\frac{3}{4}}\] Now, multiply by the speed of light, \(c = 3.00 \times 10^8 m/s\): \[v = (3.00 \times 10^8 m/s)\sqrt{\frac{3}{4}}\] \[v \approx 2.60 \times 10^8 m/s\] The relative velocity corresponding to \(\gamma=2.00\) is approximately \(2.60 \times 10^8 m/s\).
03

Calculate the relative velocity for \(\gamma=10.0\)

Now let's plug in the Lorentz factor of 10.0: \[v = c\sqrt{1-\frac{1}{(10.0)^2}}\] Calculate the value inside the square root: \[v = c\sqrt{1-\frac{1}{100}}\] \[v = c\sqrt{\frac{99}{100}}\] Finally, multiply by the speed of light, \(c = 3.00 \times 10^8 m/s\): \[v = (3.00 \times 10^8 m/s)\sqrt{\frac{99}{100}}\] \[v \approx 2.99 \times 10^8 m/s\] The relative velocity corresponding to \(\gamma=10.0\) is approximately \(2.99 \times 10^8 m/s\).

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