91Ó°ÊÓ

Consider a length of plane from observer is$l=\left(50−1.0â‹\dots {10}^{−10}\right)\text{m}$.

Consider a length of plane at rest is $l_0=50\text{m}$.

Write the formula of derived expression for the travel at a speed.

$\mathrm{Ï\dots }=\mathbf{c}\sqrt{\mathbf{1}−\frac{{\mathrm{ι}}^2}{{\mathrm{ι}}_0^2}}$ …… (1)

Here, $\mathbf{c}$ speed of light, $\mathrm{ι}$ is length contraction of an object.

Thefollowingequationmaybeusedtoshowhowarelativisticeffectcausesanobject'slengthtocontract:

$\mathrm{ι}={\mathrm{ι}}_0\sqrt{1−\frac{{\mathrm{Ï\dots }}^2}{c^2}}$

Arrive at the following formula for using the length contraction equation:

$\begin{array}{l}\mathrm{ι}={\mathrm{ι}}_0\sqrt{1−\frac{{\mathrm{Ï\dots }}^2}{c^2}}\\ \frac{\mathrm{ι}}{{\mathrm{ι}}_0}=\sqrt{1−\frac{{\mathrm{Ï\dots }}^2}{c^2}}\\ \frac{{\mathrm{ι}}^2}{{\mathrm{ι}}_0^2}=1−\frac{{\mathrm{Ï\dots }}^2}{c^2}\\ \frac{{\mathrm{Ï\dots }}^2}{c^2}=1−\frac{{\mathrm{ι}}^2}{{\mathrm{ι}}_0^2}\end{array}$

Solve further as

$\begin{array}{l}\frac{\mathrm{Ï\dots }}{c}=\sqrt{1−\frac{{\mathrm{ι}}^2}{{\mathrm{ι}}_0^2}}\\ \mathrm{Ï\dots }=c\sqrt{1−\frac{{\mathrm{ι}}^2}{{\mathrm{ι}}_0^2}}\end{array}$

Determine for using the speed's derived expression:

Substitute for$\left(3×{10}^8\right)$, $c$ $\left(50−1.0â‹\dots {10}^{−10}\right)$for role="math" localid="1659321081955" ${\mathrm{ι}}^2$ and ${50}^2$ for ${\mathrm{ι}}_0^2$ into equation (1).

$\begin{array}{c}\mathrm{Ï\dots }=3×{10}^8\sqrt{1−\frac{{\left(50−1.0â‹\dots {10}^{−10}\right)}^2}{{50}^2}}\\ =\left(2â‹\dots {10}^{−6}\right)\left(3×{10}^8\right)\\ ≈600\frac{\text{m}}{\text{s}}\end{array}$

The plane has to move at a speed of $600\frac{\text{m}}{\text{s}}$ for its length to seem $0.10\text{ n³¾}$ shorter from the observer's perspective.