How fast must be a plane 50 m long travel to be found by observer on the ground to be 0.10 nm shorter than 50 m?

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

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.

$\upsilon =\mathbf{c}\sqrt{\mathbf{1}-\frac{{\iota }^2}{{\iota }_0^2}}$ …… (1)

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

Thefollowingequationmaybeusedtoshowhowarelativisticeffectcausesanobject'slengthtocontract:

$\iota ={\iota }_0\sqrt{1-\frac{{\upsilon }^2}{c^2}}$

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

$\begin{array}{l}\iota ={\iota }_0\sqrt{1-\frac{{\upsilon }^2}{c^2}}\\ \frac{\iota }{{\iota }_0}=\sqrt{1-\frac{{\upsilon }^2}{c^2}}\\ \frac{{\iota }^2}{{\iota }_0^2}=1-\frac{{\upsilon }^2}{c^2}\\ \frac{{\upsilon }^2}{c^2}=1-\frac{{\iota }^2}{{\iota }_0^2}\end{array}$

Solve further as

$\begin{array}{l}\frac{\upsilon }{c}=\sqrt{1-\frac{{\iota }^2}{{\iota }_0^2}}\\ \upsilon =c\sqrt{1-\frac{{\iota }^2}{{\iota }_0^2}}\end{array}$

Determine for using the speed's derived expression:

Substitute for$\left(3\times {10}^8\right)$, $c$ $\left(50-1.0\cdot {10}^{-10}\right)$for role="math" localid="1659321081955" ${\iota }^2$ and ${50}^2$ for ${\iota }_0^2$ into equation (1).

$\begin{array}{c}\upsilon =3\times {10}^8\sqrt{1-\frac{{\left(50-1.0\cdot {10}^{-10}\right)}^2}{{50}^2}}\\ =\left(2\cdot {10}^{-6}\right)\left(3\times {10}^8\right)\\ \approx 600\frac{\text{m}}{\text{s}}\end{array}$

The plane has to move at a speed of $600\frac{\text{m}}{\text{s}}$ for its length to seem $0.10\text{\hspace{0.33em}nm}$ shorter from the observer's perspective.