Given that \(g\), the acceleration of gravity at the earth's surface, is \(\sim 9.8 \mathrm{~m} \mathrm{~s}^{-2}\), and that a year has \(\sim 3.2 \times 10^{7}\) seconds, verify that, in units of years and light years, \(g \approx 1\). A rocket moves from rest in an inertial frame \(S\) with constant proper acceleration \(g\) (thus giving maximum comfort to its passengers). Find its Lorentz factor relative to \(\mathrm{S}\) when its own clock indicates times \(\tau=1\) day, 1 year, 10 years. Find also the corresponding distances and times travelled in S. If the rocket accelerates for 10 years of its own time, then decelerates for 10 years, and then repeats the whole manoeuvre in the reverse direction, what is the total time elapsed in \(\mathrm{S}\) during the rocket's absence?

Step by step solution

01

Verify statement about g

To verify that \(g \approx 1\) when expressed in units of years and light years, we have to convert the given g from meters per second squared to light years per year squared. We'll use the conversion factors: 1 year has about \(3.2 \times 10^7\) seconds and 1 light year is around \(9.461 \times 10^{15}\) meters. $$ g' = \frac{9.8 \frac{\text{m}}{\text{s}^2}}{1 \frac{\text{ly}}{\text{y}^2}} = \frac{9.8 \frac{\text{m}}{\text{s}^2}}{\frac{9.461 \times 10^{15} \text{m}}{(3.2 \times 10^7 \text{s})^2}} $$ Calculate the value of \(g'\): $$ g' \approx 1 $$ Thus, the statement is verified.

02

Find Lorentz factor, distances, and times

We have three scenarios where \(\tau\) = 1 day, 1 year, and 10 years. The proper distance traveled by the rocket is given by the equation: $$ Δx' = \frac{c^2}{g} (\sqrt{1 + (gτ/c)^2} - 1) $$ And the proper time elapsed is given by: $$ Δt = \frac{c}{g} (\sqrt{1 + (gτ/c)^2} - 1) $$ When calculating distances and times, we can make use of the fact that g is approximately equal to 1 in the required units. The Lorentz factor is given by the equation: $$ γ = \sqrt{1 + (gτ/c)^2} $$ Let's find out the values for each scenario: 1. 𝜏 = 1 day 2. 𝜏 = 1 year 3. 𝜏 = 10 years

03

Calculate total elapsed time for the round trip

The forward journey consists of two equal parts: accelerating for 10 years of proper time (t1) and decelerating for 10 years of proper time (t2). The same applies to the return journey (t3 and t4). To find the total time elapsed in S, we have to sum up the time elapsed in each part: $$ Δt_\text{total} = Δt_1 + Δt_2 + Δt_3 + Δt_4 $$ Using the previously calculated proper time formula (with the duration being 10 years) and considering that the time elapsed during acceleration and deceleration is equal: $$ Δt_\text{total} = 2 (Δt_{\text{forward}} + Δt_{\text{return}}) $$ After computing the total time elapsed, we can state the final answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!