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 planet and 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 transformation equations 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`.