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Chapter 37: Q7P (page 1145)


The premise of the Planet of the Apes movies and books is that hibernating astronauts travel far into Earth鈥檚 future, to a time when human civilization has been replaced by an ape civilization. Considering only special relativity, determine how far into Earth鈥檚 future the astronauts would travel if they slept for 120 y while traveling relative to Earth with a speed of 0.9990c, first outward from Earth and then back again?

Short Answer

Expert verified

The duration for future travel by astronaut is 2.683脳103y.

Step by step solution

01

Identification of given data

The speed of astronauts relative to Earth is v=0.9990c

The duration of the sleep of an astronaut is t=120y

The time dilation is used to find the duration of particles before decaying from the detector.

02

Determination of duration for future travel by astronaut

The duration for future travel by astronaut is given as:

鈭�t=t1-vc2

Here, is the speed of light and its value is 3脳108m/s .

Substitute all the values in equation.

鈭�t=120y1-0.9990cc2鈭�t=2.683脳103y

Therefore, the duration for future travel by astronaut is 2.683脳103y.

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Most popular questions from this chapter

The car-in-the-garage problem. Carman has just purchased the world鈥檚 longest stretch limo, which has a proper length of Lc=30.5鈥尘. In Fig. 37-32a, it is shown parked in front of a garage with a proper length of Lg=6.00鈥尘. The garage has a front door (shown open) and a back door (shown closed).The limo is obviously longer than the garage. Still, Garageman, who owns the garage and knows something about relativistic length contraction, makes a bet with Carman that the limo can fit in the garage with both doors closed. Carman, who dropped his physics course before reaching special relativity, says such a thing, even in principle, is impossible.

To analyze Garageman鈥檚 scheme, an xc axis is attached to the limo, with xc=0 at the rear bumper, and an xg axis is attached to the garage, with xg=0 at the (now open) front door. Then Carman is to drive the limo directly toward the front door at a velocity of 0.9980c(which is, of course, both technically and financially impossible). Carman is stationary in the xcreference frame; Garageman is stationary in the role="math" localid="1663064422721" Xgreference frame.

There are two events to consider. Event 1: When the rear bumper clears the front door, the front door is closed. Let the time of this event be zero to both Carman and Garageman: tg1=tc1=0. The event occurs at xg=xc=0. Figure 37-32b shows event 1 according to the xg reference frame. Event 2: When the front bumper reaches the back door, that door opens. Figure 37-32c shows event 2 according to the xg reference frame.

According to Garageman, (a) what is the length of the limo, and what are the spacetime coordinates (b) xg2 and (c) tg2 of event 2? (d) For how long is the limo temporarily 鈥渢rapped鈥� inside the garage with both doors shut? Now consider the situation from the xc reference frame, in which the garage comes racing past the limo at a velocity of 鈭�0.9980c. According to Carman, (e) what is the length of the passing garage, what are the spacetime coordinates (f) Xc2and (g) tc2 of event 2, (h) is the limo ever in the garage with both doors shut, and (i) which event occurs first? (j) Sketch events 1 and 2 as seen by Carman. (k) Are the events causally related; that is, does one of them cause the other? (l) Finally, who wins the bet?

Relativistic reversal of events. Figures 37-25a and b show the (usual) situation in which a primed reference frame passes an unprimed reference frame, in the common positive direction of the xand xaxes, at a constant relative velocity of magnitude . We are at rest in the unprimed frame; Bullwinkle, an astute student of relativity in spite of his cartoon upbringing, is at rest in the primed frame. The figures also indicate events Aand Bthat occur at the following spacetime coordinates as measured in our unprimed frame and in Bullwinkle鈥檚 primed frame:

Event Unprimed Primed

AB (xA,tA)(xB,tB) (x'A,t'A)(x'B,t'B)

In our frame, event Aoccurs before event B, with temporal separation鈭�t=tA-tB=1.00渭sand spatial separation鈭�x=xB-xA=400m. Let be the temporal separation of the events according to Bullwinkle. (a) Find an expression for鈭�t'in terms of the speed parameter 尾=vcand the given data. Graph鈭�t'versus 尾for the following two ranges of尾:

(b) 0to0.01( vis low, from 0 to0.01c)

(c) 0.1 to 1 ( vis high, from to the limitc)

(d) At what value of 尾is 鈭�t'=0? For what range of 尾is the sequence of events Aand Baccording to Bullwinkle (e) the same as ours and (f) the reverse of ours? (g) Can event Acause event B, or vice versa? Explain

An unstable high-energy particle enters a detector and leaves a track of length 1.05 mm before it decays. Its speed relative to the detector was 0.992c. What is its proper lifetime? That is, how long would the particle have lasted before decay had it been at rest with respect to the detector?

Figure 37-17 shows two clocks in stationary frame S'(they are synchronized in that frame) and one clock in moving frame S. Clocks C1and C'1read zero when they pass each other. When clocks C1and C'2pass each other, (a) which clock has the smaller reading and (b) which clock measures a proper time?

Question: Temporal separation between two events. Events and occur with the following spacetime coordinates in the reference frames of Fig. 37-25: according to the unprimed frame,andaccording to the primed frame, (xA,tA)and(xB,tB). In the unprimed frame 鈭�t=tB-tA=1.00渭sand 螖x=xB-xA=240m. (a) Find an expression for 螖t'in terms of the speed parameter尾and the given data. Graph 螖t' versus 尾for the following two ranges of 尾: (b) 0 to 0.01and 0.1 (c) 0.1 to 1. (d) At what value of 尾is 螖t'minimum and (e) what is that minimum? (f) Can one of these events cause the other? Explain.
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