{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

33_InstSolManual_PC

# 33_InstSolManual_PC - RELATIVITY E XERCISES Section 33.2...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: RELATIVITY E XERCISES Section 33.2 Matter, Motion, and Ether 13. I NTERPRET In this problem we are asked to take wind speed into consideration to calculate the travel time of an airplane. D EVELOP Since the velocities are small compared to c , we can use the non-relativistic Galilean transformation of velocities in Equation 3.7, , u u v ′ = + r r r where u r is the velocity relative to the ground ( S ), u ′ r is that relative to the air ( ), S ′ and v is that of S ′ relative to S (in this case, the wind velocity). We used a notation consistent with that in Equations 33.5a and 33.5b. E VALUATE (a) If v = r (no wind), then u u ′ = r r (ground speed equals air speed), and the round-trip travel time is 2 2(1800 km) 4.5 h 800 km/h a d t u = = = (b) If v r is perpendicular to , u r then 2 2 2 , u u v ′ = + or 2 2 2 2 (800 km/h) (130 km/h) 789 km/h u u v ′ =- =- = and the round-trip travel time is 2 / 4.56 h. b t d u = = (c) If v r is parallel or anti-parallel to u r on alternate legs of the round-trip, then u u v ′ = ± and the travel time is (see Equation 33.2, but with c replaced by ) u ′ 1800 km 1800 km 4.62 h 800 km/h 130 km/h 800 km/h 130 km/h c d d t u v u v = + = + = ′ ′ +- +- A SSESS We find , a b c t t t < < as mentioned in the paragraph following Equation 33.2. 14. The difference between Equations 33.2 and 33.1 is || 2 2 2 2 2 2 2 2 2 2 2 1 1 1 / 1 / cL L L t t t c c v v c c v v c ⊥ ∆ =- =- =- ---- and 8 2 / 2(11 m)/(3 10 m/s) 73.3 ns L c = × = for the given interferometer. If 2 2 / 1, v c << we can expand the denominators (Appendix A) to obtain 2 2 2 2 2 2 (2 / )[1 / (1 /2 )] (2 / )( /2 ). t L c v c v c L c v c ∆ ≈ +- + = (a) For the Earth’s orbital speed (30 km/s), 2 2 8 / 10 , v c- = so that 8 16 1 2 73.3 ns( 10 ) 3.67 10 s t-- ∆ = × = × (a fraction of the period of visible light). (b) If 2 2 4 12 / 10 , 3.67 10 s v c t-- = ∆ = × (a few thousand periods of visible light). (c) At relativistic speeds, we use the exact expression for . t ∆ If 2 2 2 2 1 4 / , 1/ 1 / 2/ 3, v c v c γ = =- = and 4 3 73.3 ns( 2/ 3) 13.1 ns. t ∆ =- = (d) If / 0.99, v c = 1/ 1 0.98 7.09, γ =- = and (2 / ) ( 1) 3.17 s. t L c γ γ μ ∆ =- = Section 33.4 Space and Time in Relativity 15. I NTERPRET This problem is about the distance between two stars, as seen by a spaceship moving at relativistic speed. D EVELOP The distance between stars at rest in system S appears Lorentz-contracted in the spaceship’s system , S ′ according to Equation 33.4: 33.1 33 33.2 Chapter 33 2 2 1 / x x v c ′ ∆ = ∆- E VALUATE With 50 ly x ∆ = and 0.75 , v c = we get 2 2 2 1 / (50 ly) 1 (0.75) 33.1 ly x x v c ′ ∆ = ∆- =- = A SSESS The distance appears to be shortened or “contracted” as observed by the spaceship. Note that length contraction occurs only along the direction of motion....
View Full Document

{[ snackBarMessage ]}

### Page1 / 14

33_InstSolManual_PC - RELATIVITY E XERCISES Section 33.2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online