To circle Earth in low orbit, a satellite must have a speed of about 2.7 x 104 km/h. Suppose that two such satellites orbit Earth in opposite directions. (a) What is their relative speed as they pass, according to the classical Galilean velocity transformation equation? (b) What fractional error do you make in (a) by not using the (correct) relativistic transformation equation?

Short Answer

Expert verified

In the classical limit, the relative speed is 5.4×104km/h and the fractional error is1.85×10-3 when compared relativistic limit.

Step by step solution

01

(a )Classical Galilean velocity transformation equation.

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'with respect to a frame S’ which in turn is moving at velocity vwith respect to a stationary frame S. Therefore, according to this Galilean relativity, the velocity role="math" localid="1663141769904" u'wrt the S frame is

u=u'+v

For the given case, both satellites are moving at the same speed of 2.7×104km/hbut moving towards each other. Therefore, the relative speed will be

u=2v=2(2.7×104km/h)=5.4×104km/h

02

(b )Relativistic velocity transformation.

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=u'+v1+u'vc2

Substituting the value of velocities

u=2v1+v2c2=2(0.75×104m/s)1+(0.75×104m/s)2(3×108m/s)2=14999.9m/s=5.39×104km/h

The fractional error is

5.39×104km/h-5.40×104km/h5.39×104km/h=1.85×10-3

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

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