Question

According to Bob on Earth, Planet Y (uninhabited) is 5 ly away. Anna is in a spaceship moving away from Earth at 0.8c. She is bound for planet Y to study its geology. Unfortunately, Planet Y explodes. According to Bob.This occurred 2 yr after Anna passed Earth. (Bob. of course. has, to wait a while for the light from the explosion to arrive, but he reaches his conclusion by “working backward”) Call the passing of Anna and Bob time zero for both. (a) According to Anna, how far away is Planet Y when it explodes? (b) At what time does it explode?

Short answer

(a) The distance of planet Y according to Anna is $5.67\mathrm{ly}$.

(b) The time of explosion is $7.07\mathrm{yr}$.

Step-by-step solution

Identification of the given data

The given data is:

• The distance of the planet Y is $\mathrm{x}=5\mathrm{ly}$.
• The speed of spaceship is $\mathrm{v}=0.8\mathrm{c}$.
• The duration for occurrence of event is $\mathrm{t}=2\mathrm{yr}$.

Analysis of concept used

The above problem is based on the concept of Lorentz transformations and time dilation. All the distances in light-year should be changed to meters for calculation.

(a) Determination of the distance of planet Y according to Anna

The formula to calculate the distance of planet Y is given as:

$\mathrm{d}=\frac{\mathrm{x}-\mathrm{vt}}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}$

Substitute all the values in the above equation,

$$\begin{gathered} \mathrm{d}=\frac{\left(5\mathrm{ly}\right)\left({\displaystyle \frac{9.461\times {10}^{15}\mathrm{m}}{1\mathrm{ly}}}\right)-\left(0.8\mathrm{c}\right)\left({\displaystyle \frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\mathrm{c}}}\right)\left(2\mathrm{yr}\right)\left({\displaystyle \frac{3.154\times {10}^7\mathrm{s}}{1\mathrm{yr}}}\right)}{\sqrt{1-{\displaystyle \frac{{\left(0.8\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{d}=53.61\times {10}^{15}\mathrm{m} \\ \mathrm{d}=\left(53.61\times {10}^{15}\mathrm{m}\right)\left(\frac{1\mathrm{ly}}{9.461\times {10}^{15}\mathrm{m}}\right) \\ \mathrm{d}=5.67\mathrm{ly} \end{gathered}$$

Therefore, the distance of planet Y according to Anna is data-custom-editor="chemistry" $5.67\mathrm{ly}$.

 (b) Determination of the time of explosion of planet Y

The time of explosion of planet Y can be given as:

$\mathrm{T}=\frac{\left(-{\displaystyle \frac{\mathrm{vx}}{{\mathrm{c}}^2}}+\mathrm{t}\right)}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}$

Substitute all the values in the above equation,

$$\begin{gathered} \mathrm{T}=\frac{-{\displaystyle \frac{\left(0.8\mathrm{c}\right)\left({\displaystyle \frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\mathrm{c}}}\right)\left(5\mathrm{ly}\right)\left({\displaystyle \frac{9.461\times {10}^{15}\mathrm{m}}{1\mathrm{ly}}}\right)}{{\left(\left(0.8\mathrm{c}\right)\left({\displaystyle \frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\mathrm{c}}}\right)\right)}^2}}+\left(2\mathrm{yr}\right)\left({\displaystyle \frac{3.154\times {10}^7\mathrm{s}}{1\mathrm{yr}}}\right)}{\sqrt{1-{\displaystyle \frac{{\left(0.8\mathrm{c}\right)}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{T}=-22.33\times {10}^7\mathrm{s} \\ \mathrm{T}=\left(-22.33\times {10}^7\mathrm{s}\right)\left(\frac{1\mathrm{yr}}{3.154\times {10}^7\mathrm{s}}\right) \\ \mathrm{T}=-7.07\mathrm{yr} \end{gathered}$$

Here, the negative sign indicates the explosion on planet Y, which occurs before explosion observed by Anna, so time is negative.

Therefore, the time of explosion is data-custom-editor="chemistry" $7.07\mathrm{yr}$.