Question

Astronaut Spud Nick is space-traveling from plnaet X to planet Y at a speed of 0.60c relative to the planets, whch are at rest relative to each other. Whne he is precisely halway between the planets, a distnace of 1.0 light-hour from each one as measured in the panet frame, nuclear devices are detonated on each planet. The explosions are simultaneous in the frame of the planets. What is the difference in the time of arrival of the glshes from the xplosions as observed by Spud?

Answer 1

Answer:

112.5 min

Explanation:

$L_0$ = Original length = 1 lh = $1c$

Length contraction

$L=L_0(\sqrt{1-\dfrac{u^2}{c^2}})$

The difference in time is given by

$\Delta t=\dfrac{L}{c-u}-\dfrac{L}{c+u}\\\Rightarrow \Delta t=\dfrac{L(c+u)-L(c-u)}{c^2-u^2}\\\Rightarrow \Delta t=\dfrac{L(c+u-c+u)}{c^2-u^2}\\\Rightarrow \Delta t=\dfrac{2Lu}{c^2-u^2}$

$L=L_0\sqrt{1-\dfrac{u^2}{c^2}}$

$\Delta t=\dfrac{2uL_0\sqrt{1-\dfrac{u^2}{c^2}}}{c^2-u^2}\\\Rightarrow \Delta t=\dfrac{2uL_0\sqrt{1-\dfrac{u^2}{c^2}}}{(1-\dfrac{u^2}{c^2})c^2}\\\Rightarrow \Delta t=\dfrac{2\times 0.6c 1c\sqrt{1-\dfrac{0.6^2c^2}{c^2}}}{(1-\dfrac{0.6^2c^2}{c^2})c^2}\\\Rightarrow \Delta t=1.5\ hr$

$1.5\times \dfrac{60}{\sqrt{1-\dfrac{0.6^2c^2}{c^2}}}=112.5\ min$

$1.5\times \dfrac{60}{\sqrt(1-0.36)}\\ =1.5 \times \dfrac{60}{\sqrt{0.64}}\\ =1.5\times \dfrac{60}{0.8}\\ =\dfrac{90}{0.8}\\ =112.5\ minutes$

The time taken is 112.5 min