Question

The radius of Earth is $\mathbf{6370}\mathit{k}\mathit{m}$, and its orbital speed about the Sun is $\mathbf{30}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. Suppose Earth moves past an observer at this speed. To the observer, by how much does Earth’s diameter contract along the direction of motion?

Short answer

In the rest frame of the observer, the diameter of earth contracts by$\mathbf{8}\mathit{m}\mathit{m}$.

Step-by-step solution

Lorentz factor:

The result of the ${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$postulate of special relativity is that the clock runs slower for a moving object when measured from a rest frame. The factor by which the clocks run differently is called the Lorentz factor.

The earth is moving at $30km/s$relative to observer’s frame. The Lorentz factor for this speed is

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ =\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(30km/s\right)}^2}{{\left(3\times {10}^{5}km/s\right)}^2}}}} \\ =\text{1.000000006} \end{gathered}$$

Length contraction:

The length of an object measured in the object’s inertial reference frame is called proper length. When the length of this object is measured in any other inertial frame moving at relative speed is always shorter than the proper length. This is known as length contraction.

$$\begin{gathered} L=L_o\sqrt{1-{\beta }^2} \\ =\frac{L_o}{\gamma } \end{gathered}$$

Here, the diameter measured from the earth’s frame (proper length) is $12,740km$, and the Lorentz factor is found to have a value of $1.000000006$.

The distance measured from the observer’s reference(stationary) frame is

$$\begin{gathered} L=\frac{L_o}{\gamma } \\ =\frac{12740km}{1.000000006} \\ =12739.99992km \end{gathered}$$

In the rest frame of the observer, the diameter of earth contracts by,

$$\begin{gathered} 12740km-12739.99992km=0.000008km \\ =8mm \end{gathered}$$