Question

Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missile's clock to fall behind the ground observer's clock by 1 millisecond? Hint: use the binomial formula:(1+x)a1+ ax.

Answer 1

Answer:

The time taken by missile's clock is $4.6\times 10^{6} s$

Solution:

As per the question:

Speed of the missile, $v_{m = 6.5\times 10^{3}} m/s$

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:

$T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}$         (1)

Using binomial theorem in the above eqn:

We know that:

$(1 + x)^{a} = 1 + ax$

Thus eqn (1) becomes:

$1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}$

$T = \frac{2c^{2}}{v_{m}^{2}}$

Now, putting appropriate values in the above eqn:

$T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}$

$T = 4.6\times 10^{6} s$