91Ó°ÊÓ

In Galilean-newton’s relativity, it is assumed that all the laws of physics are the same irrespective of the inertial frame of reference. The space and time intervals are considered to be absolute.

Suppose an object is moving at velocity $u\text{'}$with respect to a frame S’ which in turn is moving at velocity $v$with respect to a stationary frame S. Therefore, according to this Galilean relativity, the velocity role="math" localid="1663141769904" $u\text{'}$wrt the S frame is

$u=u\text{'}+v$

For the given case, both satellites are moving at the same speed of $2.7×{10}^{4}km/h$but moving towards each other. Therefore, the relative speed will be

$$\begin{gathered} u=2v \\ =2(2.7×{10}^{4}km/h) \\ =5.4×{10}^{4}km/h \end{gathered}$$

Einstein proposed that all the basic laws of physics behave the same in all inertial frames of reference. And the speed of light is a constant independent of which reference frame. As a result, the classical velocity addition equation cannot be used for relativistic speeds. The relative velocities for relativistic speeds are calculated using the following equation.

$u=\frac{u\text{'}+v}{1+\frac{u\text{'}v}{c^2}}$

Substituting the value of velocities

$$\begin{gathered} u=\frac{2v}{1+{\displaystyle \frac{v^2}{c^2}}} \\ =\frac{2(0.75×{10}^{4}m/s)}{1+{\displaystyle \frac{{(0.75×{10}^{4}m/s)}^2}{{(3×{10}^{8}m/s)}^2}}} \\ =14999.9m/s \\ =5.39×{10}^{4}km/h \end{gathered}$$

The fractional error is

$\frac{5.39×{10}^{4}km/h-5.40×{10}^{4}km/h}{5.39×{10}^{4}km/h}=1.85×{10}^{-3}$

The error is minuscule because the velocity of the satellite is quite small compared to the speed of light.