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Chapter 35: Problem 22

Find the value of \(g\), the gravitational acceleration at Earth's surface, in light-years per year, to three significant figures.

Short Answer

Expert verified
Answer: 1.13 x 10^-16 ly/yr²

Step by step solution

01

Determining Known Values and Conversion Factors

The known values required are: - \(g\) (Earth's gravitational acceleration) in meters per second squared (m/s^2): \(g = 9.81\,\text{m/s}^2\). - \(c\) (speed of light) in meters per second (m/s): \(c = 3.00\times10^8\,\text{m/s}\). - \(t\) (number of seconds in a year): \(t = 3.15\times10^7\,\text{s}\). With these known values, we can come up with the necessary conversion factors: - Meters to light-years: \(1\,\text{light-year} = \frac{c \times t}{1\,\text{year}}\). - Seconds to years: \(1\,\text{year} = 3.15 \times 10^7\,\text{s}\). Step 2:
02

Converting Earth's Gravitational Acceleration to Light-Years Per Year

We will first find the conversion factor for meters to light-years: \(1\,\text{light-year} = \frac{c \times t}{1\,\text{year}} = \frac{3.00\times10^8\,\text{m/s} \times 3.15\times10^7\,\text{s}}{1\,\text{year}} = 9.46\times10^{15}\,\text{m}\). Now, we can convert the value of \(g\) from meters per second squared to light-years per year squared: \(g_{\text{ly/yr}^2} = \left(\frac{9.81\,\text{m/s}^2}{9.46\times10^{15}\,\text{m/ly}}\right)\times\left(\frac{1\,\text{year}}{3.15\times10^7\,\text{s}}\right)^2\) Step 3:
03

Rounding the Result to Three Significant Figures

Evaluating the previous expression, we get: \(g_{\text{ly/yr}^2} \approx 1.1251 \times 10^{-16}\,\text{ly/yr}^2\) To round the result of \(g\) to three significant figures, we get: \(g_{\text{ly/yr}^2} \approx 1.13 \times 10^{-16}\,\text{ly/yr}^2\) So, the value of Earth's gravitational acceleration in light-years per year squared, to three significant figures, is \(1.13 \times 10^{-16}\,\text{ly/yr}^2\).

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Most popular questions from this chapter

Prove that in all cases, two sub-light-speed velocities "added" relativistically will always yield a sub-light-speed velocity. Consider motion in one spatial dimension only.

Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means \(d v / d t=g,\) where \(g\) is a constant, since that would have \(v\) exceeding \(c\) in a finite time. Rather, it means that the acceleration experienced by the moving body is constant: In each increment of the body's own proper time \(d \tau,\) the body acquires velocity increment \(d v=g d \tau\) as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation: a) Write a differential equation for the velocity \(v\) of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation, remembering that squares and higher powers of differentials can be neglected. b) Solve this equation for \(v(t),\) where both \(v\) and \(t\) are measured in the ground frame. c) Verify that your solution behaves appropriately for small and large values of \(t\). d) Calculate the position of the body \(x(t),\) as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time \(t=0,\) at ground-frame position \(x=c^{2} / g\) e) Identify the trajectory of the body on a space-time diagram (Minkowski diagram, for Hermann Minkowski) with coordinates \(x\) and \(c t,\) as measured in the ground frame. f) For \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) calculate how much time it takes the body to accelerate from rest to \(70.7 \%\) of \(c,\) measured in the ground frame, and how much ground-frame distance the body covers in this time.

At rest, a rocket has an overall length of \(L .\) A garage at rest (built for the rocket by the lowest bidder) is only \(L / 2\) in length. Luckily, the garage has both a front door and a back door, so that when the rocket flies at a speed of \(v=0.866 c\), the rocket fits entirely into the garage. However, according to the rocket pilot, the rocket has length \(L\) and the garage has length \(L / 4\). How does the rocket pilot observe that the rocket does not fit into the garage?

In the twin paradox example, Alice boards a spaceship that flies to a space station 3.25 light-years away and then returns with a speed of \(0.650 c .\) This can be viewed in terms of Alice's reference frame. a) Show that Alice must travel with a speed of \(0.914 c\) to establish a relative speed of \(0.650 c\) with respect to Earth when Alice is returning back to Earth. b) Calculate the time duration for Alice's return flight toward Earth with the aforementioned speed.

If a muon is moving at \(90.0 \%\) of the speed of light, how does its measured lifetime compare to when it is in the rest frame of a laboratory, where its lifetime is \(2.2 \cdot 10^{-6}\) s?
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