InterviewSolution
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InterviewSolution
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An Earth starship has been sent to check an Earth outpost on the planet P1407, whose moon houses a battle group of the often hostile Reptulians. As the ship follows a straightline course first past the planet and then past the moon, it detects a high-energy microwaveburst at the Reptulian moon base and then, 1.10 s later, an explosion at the Earth outpost, which is 4.00xx10^(8) m from the Reptulian base as measured from the ship's reference frame. The Reptulians have obviously attacked the Earth outpost, and so the starship begins to prepare for a confrontation with them. The speed of the ship relative to the planet and its moon is 0.980c. What are the distanceand time interval between the burst and the explosion as measured in the planet-moon frame (and thus according to the occupants of the stations)? |
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Answer» SOLUTION :KEY IDEAS 1. This problem involves measurements made from two reference frames, the planet-moon frame and the starship frame. 2.We have two EVENTS: the burst and the explosion. 3. We need to TRANSFORM the given data as measured in the starship frame to the corresponding data as measured in the planet-moon frame. Starship frame: Before we get to the transformation, we need to carefully choose our notation. We begin with a sketch of the situation as shown in Fig. 36-12. There, we have chosen the ship.s frame S to be stationary and the planet-moon frame S. to be moving with positive velocity (rightward). (This is an arbitrary choice, we could, instead, have chosen the planet-moon frame to be stationary. Then we WOULD redraw `vecv` in Fig. 36-12 as being attached to the S frame and indicating leftward motion, v would then be a negative quantitiy. The results would be the same.) Let subscripts e and b represent the explosion and burst, respectively. Then the given darta, all in the unprimed (starship) reference frame, are `Deltax=x_(e )-x_(b)= +4.00xx10^(8)m` and `Deltat=t_(e )-t_(b)=+1.10s`. The relative motion alters the time intervals between events and maybe even their sequence.  Figure35-12 A planetand its moon in reference frame S. move rightward with speed v relative to a starship in reference frame S. Here, Deltax is a positive quantity becasuse in Fig. 36-12, the coordinate `x_(e )` for the explosion in greater than the coordinate `x_(b)` for the burst, `Deltat` is also a positive quantity because the time `t_(e )` of the explosion is greater (later) than the time `t_(b)` of the burst. Planet-moon frame: We seek `Deltax.` and `Deltat.`, which we shall get by transforming the given S-frame data to the planet-moon frame S.. Because we are considering a pair to events, we choose transformationequations from Table 36-2-namely, EQS. 1. and2.: M `Deltax.=gamma (Deltax-v Deltat)""`(36-27) and `Deltat.=gamma(Delta-(v Deltax)/(c^(2))).""`(36-28) Here, `v = +980c` and the Lorentz factor is `gamma=(1)/(sqrt(1-(v//c)^(2)))=(1)/(sqrt(1-(+0.980c//c)^(2)))=5.0252.` Equation 36-27 then becomes `Deltax.=(5.0252)[4.00xx10^(8)m-(+0.980c)(1.10s)]` `=3.86xx10^(8)m`, and Eq 36-28 becomes `Deltat.=(5.0252)[(1.10s)-((+0.980c)(4.00xx10^(8)m))/(c^(2))]` `= -1.04s`. |
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