Question

Assuming that the rest radius of earth is 6400 km and its orbital speed about the sun is 30 km/sec,

how much does earths diameter appear to be shortened to an observer on the sun, due to earth’s

orbital motion?

Answer 1

Answer

Radius of earth 6400Km

Speed v=30km/sec

$β =\frac{ v}{c}\\ = \frac{30 km s^{−1}}{3 × 10^5km^{ s−1 }}\\ = 10^{−4}$

Now

$\frac{1}{γ} = (1 − β^2)^{1/2 }\\ = (1 − \frac{1}{2} × β^2)$

So earth shrunk

$ΔL = L_0 − L \\ = L_0 −L_0/γ \\ = L_0 − L_0(1 − β^2)^{\frac{1}{2} }\\ = \frac{1}{2}L_0β^2 \\ = \frac{1}{2} ×6,400× 10^{−8} km\\ = 3.2 cm$

Thus the earth appears to be shrunk by 3.2 cm