Problem 71 - Fundamentals of Physics

Question

To circle Earth in low orbit, a satellite must have a speed of about \(2.7 \times 10^{4} \mathrm{~km} / \mathrm{h}\). 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?

Step-by-Step Solution

Answer

(a) 54000 km/h; (b) Fractional error is 0.

Step 1: Calculate Relative Speed using Galilean Transformation

According to the classical Galilean velocity transformation, the relative speed between two objects moving towards each other is simply the sum of their speeds. Each satellite moves with a speed of \( 2.7 \times 10^4 \) km/h relative to the Earth. Therefore, the relative speed \( v_{\text{relative}} \) is given by:\[ v_{\text{relative}} = v_1 + v_2 = 2.7 \times 10^4 + 2.7 \times 10^4 = 5.4 \times 10^4 \text{ km/h} \]

Step 2: Calculate Relativistic Relative Speed

According to the relativistic velocity transformation, the relative speed \( v_{\text{rel}} \) is calculated using:\[ v_{\text{rel}} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} \]where \( c \) is the speed of light in km/h, \( c \approx 1.08 \times 10^9 \text{ km/h} \). Substituting the given values:\[ v_{\text{rel}} = \frac{2.7 \times 10^4 + 2.7 \times 10^4}{1 + \frac{(2.7 \times 10^4)(2.7 \times 10^4)}{(1.08 \times 10^9)^2}} \approx 5.4 \times 10^4 \text{ km/h} \]

Step 3: Calculate Fractional Error

The fractional error is the difference between the classical and relativistic calculations divided by the relativistic calculation. Since the relativistic result is approximately the same as the classical one, the fractional error is:\[ \text{Fractional Error} = \frac{5.4 \times 10^4 - 5.4 \times 10^4}{5.4 \times 10^4} = 0 \]

Step 4: Final Conclusion

The classical calculation gives the same result as the relativistic calculation for this speed, leading to a fractional error of almost zero because the speeds involved are very small compared to the speed of light.

Key Concepts

Galilean Velocity Transformation

The Galilean Velocity Transformation is a method in classical physics used to find the relative velocity between two objects. This method assumes that velocities simply add up, without the influence of relativistic effects.
For example, if you have a car moving at 50 km/h and another one at 40 km/h in the opposite direction, their relative speed is 90 km/h. This approach is straightforward but only accurate at speeds much slower than the speed of light.
In the exercise's context, each satellite moves at a speed of \(2.7 \times 10^4\) km/h. The Galilean equation calculates their relative speed as simply the sum of their speeds:

• Relative speed = Sum of individual speeds
• \(v_{\text{relative}} = 2.7 \times 10^4 + 2.7 \times 10^4 = 5.4 \times 10^4\) km/h.

When velocities become a significant fraction of the speed of light, as is subtly the case here, relativistic effects become important, although for typical Earth-bound speeds, this method performs well.

Fractional Error

The Fractional Error is a measure of how much the Galilean transformation deviates from the more accurate relativistic calculation. It's an important concept in evaluating the precision needed when calculating velocities that are a smaller fraction of the speed of light.
To calculate the fractional error, you use the difference between the predicted classical speed and the accurate relativistic speed divided by the relativistic speed itself. In formula terms:

• \(\text{Fractional Error} = \frac{\text{Classical Speed} - \text{Relativistic Speed}}{\text{Relativistic Speed}}\)

For the orbiting satellites, both the classical and relativistic calculations produce approximately the same result: \(5.4 \times 10^4\) km/h. This leads to a fractional error very close to zero, illustrating that at speeds of this magnitude, Galilean and relativistic results align closely. However, as velocities approach a non-negligible fraction of the speed of light, this error can increase significantly.

Low Earth Orbit

Low Earth Orbit (LEO) is a term used to describe a certain range of altitudes where satellites orbit the Earth. Generally, this range extends from about 160 km to 2,000 km above the Earth's surface. Satellites here move around Earth at tremendous speeds to stay in orbit due to the planet's gravitational pull.
In the context of our exercise, satellite speed is given as \(2.7 \times 10^4\) km/h, which is necessary to counteract the pull of gravity and maintain a stable orbit. At these altitudes and speeds, communication satellites and the International Space Station operate effectively.
Characteristics of LEO include:

• Short orbital periods, usually around 90 minutes.
• Relatively close proximity to Earth, allowing low-latency communication systems.
• Re-entry decay over time without propulsion adjustment, due to atmospheric drag.

The speeds involved in LEO are high but still allow for using classical physics under most circumstances. When examined more closely at higher velocities, relativistic physics could become significant.