91Ó°ÊÓ

Consider a time difference is$\mathrm{Δ³Ù}=10\text{ s}$.

Consider a speed of the plane is$\mathrm{Ï\dots }=420\frac{\text{m}}{\text{s}}$.

Consider a distance is $\text{4200 m}$.

Write the formula of time difference in the plane’s clock.

${\mathrm{Δ³Ù}}_0=\mathrm{Δ³Ù}\sqrt{\mathbf{1}−\frac{{\mathrm{Ï\dots }}^2}{{\mathbf{c}}^2}}$ …… (1)

Here, $\mathrm{Δ}\mathbf{t}$ is time difference, $\mathrm{Ï\dots }$ speed of the plane and $\mathbf{c}$ speed of light.

Due to relativistic effects, time relative to an object moves more quickly when it is moving very quickly. The time-dilation equation helps explain this:

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

Arrive at the following expression for the time difference in the plane's clock using the time-dilation equation:

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

Calculate the time difference in the clock of the aircraft using the formula for${\mathrm{Δ³Ù}}_0$:

Determine the time difference in the plane’s clock.

Substitute $10$ for$\mathrm{Δ}t$, $420$ for $\mathrm{Ï\dots }$ and $\left(3×{10}^8\right)$ for $c^2$ into equation (1).

$\begin{array}{c}\mathrm{Δ}{t}_0=10\sqrt{1−\frac{{\left(420\right)}^2}{{\left(3×{10}^8\right)}^2}}\\ =9.8×{10}^{−12}\\ =9.8\text{ p²õ}\end{array}$

According to the findings, the plane's clock would display the time $\text{9.8 p²õ}$ earlier.