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) 5.4 × 10⁴ km/h; (b) Fractional error is ~5.4 × 10⁻⁸.

Step 1: Understanding Relative Speed with Galilean Transformation

According to classical mechanics, the relative velocity of two objects moving directly towards each other is simply the sum of their respective velocities. So, if one satellite moves with a velocity \( v \) and the other with velocity \( -v \), their relative speed is the sum: \( v + v = 2v \).

Step 2: Calculating Satellite Speed

Given that the speed of one satellite is \( v = 2.7 \times 10^4 \) km/h, applying the Galilean transformation, the relative speed \( v_{\text{relative}} \) when they pass each other is \( 2v \).

Step 3: Applying the Values

Substitute \( v = 2.7 \times 10^4 \) km/h into the formula to find the relative speed: \( v_{\text{relative}} = 2 \times 2.7 \times 10^4 = 5.4 \times 10^4 \) km/h.

Step 4: Calculating Speed Using Relativistic Transformation

Using the relativistic velocity transformation: \[ u' = \frac{u + v}{1 + \frac{uv}{c^2}} \] where \( u = 2.7 \times 10^4 \) km/h, \( v = 2.7 \times 10^4 \) km/h, and \( c = 3 \times 10^8 \) m/s (convert \( c \) to km/h). First calculate:\[ \frac{uv}{c^2} \approx \frac{(2.7 \times 10^4)^2}{(3 \times 10^5)^2} \approx 2.7 \times 10^{-8} \] Then, calculate the relativistic relative speed.

Step 5: Calculate Fractional Error

The fractional error is calculated by taking the absolute difference between the classical and relativistic speeds divided by the relativistic speed. Calculate it as: \[ \text{fractional error} = \frac{|5.4 \times 10^4 - 5.4 \times 10^4|}{5.4 \times 10^4} = 2 \times 2.7 \times 10^{-8} \]

Key Concepts

Galilean Transformation

In classical mechanics, we often encounter the concept of Galilean transformation. This transformation is used to relate the coordinates and time of an event as observed from two different inertial frames that are in uniform relative motion. It's quite straightforward and easy to digest: imagine you're on a moving train, and you want to know how fast you're moving relative to someone standing still on the platform.
Galilean transformations assume that time is absolute, meaning it flows the same way regardless of the observer's state of motion. The transformation equations are simple: if two objects are moving towards each other, their relative speed is simply the sum of their individual speeds.

• For example, if one satellite is moving at a speed of \( v \) and another at the opposite direction with speed \( -v \), the relative speed determined by Galilean transformation would be \( v + v = 2v \).

This concept works quite well at everyday speeds and is useful for solving many problems in classical mechanics, like calculating the relative speed of satellites orbiting Earth.

Relative Velocity

Relative velocity is essentially the velocity of an object as observed from a particular frame of reference. It tells us how fast or slow something is moving relative to another object. This concept is crucial for understanding how different observers perceive motion differently.
Imagine you're in a car moving at 60 km/h and someone else is moving in a different car at 40 km/h in the same direction. From your perspective, the other car appears to be moving at a speed of 20 km/h. Hence, relative velocity considers both your speed and the speed of the other object.

• The formula is simple: if you are calculating the relative speed of two objects moving in opposite directions, you simply add their speeds.

This highlights the importance of reference frames in analyzing motion, as different observers can report different velocities for the same moving object based on their unique point of view.

Relativistic Transformation

When dealing with speeds close to the speed of light, classical mechanics falls short, and we need to use concepts from Einstein's Theory of Relativity. Relativistic transformation corrects the limitations of Galilean transformations, allowing us to accurately calculate velocities when objects are moving at significant fractions of the speed of light.
The relativistic velocity transformation formula is given by:

\[ u' = \frac{u + v}{1 + \frac{uv}{c^2}} \]
where \( u \) and \( v \) are the velocities of the objects, and \( c \) is the speed of light.
Relativistic transformations account for the fact that velocities don't simply add up like they do in Galilean transformations. Instead, they factor in the effects of time dilation and length contraction.

• For the satellites circling the Earth, despite their high speed, the relativistic effects are minimal, given that their velocities are far from the speed of light. This is why the fractional error calculated is very small.

Understanding these transformations is crucial when working with high-speed phenomena.

Satellite Orbit

Satellites orbiting the Earth do so under the influence of gravity, which keeps them in a stable path around our planet. These orbits can be low, medium, or high, depending on how far the satellite is from the Earth's surface.
In a low Earth orbit (LEO), satellites travel at extremely high speeds, approximately \(2.7 \times 10^{4} \text{ km/h}\), to counterbalance the gravitational pull and maintain orbit.
When satellites orbit in opposite directions, their speeds relative to each other become significant, especially when calculating scenarios like collision avoidance.

• The process of determining relative velocity, either using Galilean transformation or relativistic transformation, becomes essential.
• These calculations ensure that the satellites avoid collisions and can operate in their orbits safely.

Understanding satellite orbits is not only vital for space travel and communication but also for advancements in technology and earth sciences.