Answer
The percentage decrease is $(6.97\times 10^{-8})\%$
Work Step by Step
Use equation 26–3a.
$$\mathcal{l} = \mathcal{l}_{o} \sqrt{1-\frac{ v^{2} } { c^{2} }} $$
$$\frac{\mathcal{l}}{\mathcal{l}_o}=(1-\frac{v^2}{c^2})^{1/2}$$
Escape velocity is much smaller the speed of light. Use the binomial expansion to find the percentage decrease.
$$\frac{\mathcal{l}}{\mathcal{l}_o}\approx 1-\frac{1}{2}\frac{v^2}{c^2}$$
$$= 1-\frac{1}{2}\frac{(11.2\times 10^3 m/s)^2}{(3.00\times 10^8 m/s)^2}$$
$$= 1-6.97\times 10^{-10}$$
The percentage decrease is $(6.97\times 10^{-8})\%$.
