Respuesta :
Answer:
The time taken by missile's clock is [tex]4.6\times 10^{6} s[/tex]
Solution:
As per the question:
Speed of the missile, [tex]v_{m = 6.5\times 10^{3}} m/s[/tex]
Now,
If 'T' be the time of the frame at rest then the dilated time as per the question is given as:
T' = T + 1
Now, using the time dilation eqn:
[tex]T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}[/tex] (1)
Using binomial theorem in the above eqn:
We know that:
[tex](1 + x)^{a} = 1 + ax[/tex]
Thus eqn (1) becomes:
[tex]1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}[/tex]
[tex]T = \frac{2c^{2}}{v_{m}^{2}}[/tex]
Now, putting appropriate values in the above eqn:
[tex]T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}[/tex]
[tex]T = 4.6\times 10^{6} s[/tex]
