## Related questions with answers

The radius of Earth is 6370 km, and its orbital speed about the Sun is 30 km/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?

Solution

VerifiedLet $D_0$ be the diameter of the earth at rest and $D$ is the diameter of the earth to the observer, the magnitude of the contraction is given by:

$\begin{align}|\Delta D|=D_0-D\end{align}$

the diameter of the earth to the observer is given by:

$\begin{align} D=\dfrac{D_0}{\gamma}\end{align}$

where $\gamma$ is the Lorentz factor and it is given by:

$\gamma=\dfrac{1}{\sqrt{1-(v/c)^2}}$

where $v$ is the radial speed. Substitute into (2) then into (1) to get:

$D=D_0\sqrt{1-(v/c)^2}$

thus,

$|\Delta D|=D_0-D_0\sqrt{1-(v/c)^2}=D_0\left(1-\sqrt{1-(v/c)^2} \right)$

substitute with the givens to get (note that $D_0=2 \times 6370 \times 10^{3} \mathrm{~m}=1.274\times 10^{7}$ m):

$\begin{align*}|\Delta D|&= (1.274\times 10^{7} \mathrm{~m})\left(1-\sqrt{1-((30 \times 10^{3} \mathrm{~m/s})/(3.0 \times 10^{8} \mathrm{~m/s}))^2} \right)\\ &=0.0637 \mathrm{~m} \end{align*}$

$\boxed{|\Delta D|=0.0637 \mathrm{~m}}$

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