An observer detects two explosions one occur near her at certain time and another occurs 1 ms later 100km away. Another observer finds the two explosions occurs at the same place. At what time interval (in ms) separates the explosions to the second observer.
Answers
Answer:
12300ms
Explanation:
because certain time and another occurs 1 ms later 100km away
Given:
An observer detects two explosions one occur near her at certain time and another occurs 1 ms later 100km away.
Explanation:
Lt the first explosion take place at an time t, in the rest and the other explosion takes place at the time t₂
Given that the time between these two explosition in 2ms
∴ t₂ - t₁ = 1ms
= 1 × 10⁻³s
Also the separation between the two points = 100km
that is , x₂ - x₁ = 100km
= 10⁵m
Let the portions of the two events for an observer moving at certain speed v are x'₁ and x'₂ also the times in that frame are t'₁ and t'₂
We have
x'₁ = x₁ -vt₁ / √ 1- (v²/c²)
x'₂ =x₂ -vt₂ / √ 1- (v²/c²)
And t'₁ = [t₁ - (vx₁/c²)] / [√ 1 - (v²/c²)]
For the observer moving with the speed v, the event are taken place at the same
∴ x'₂ - x'₁ = 0
x₂ -vt₂ / √ 1- (v²/c²) -x₁ - vt₁ / √ 1- (v²/c²) = 0
(x₂ - x₁) - v (t₂ - t₁) = 0
v = (x₂ - x₁)/ (t₂ - t₁)
= 10⁵m / 1× 10⁻³s
= 1× 10⁸ m/s
Then the time interval between two explosions to the second observer
t'₂ - t'₁ = [t₂ - (vx₂/c₂) /√ 1 -(v²/c²)] - [t₁ - vx₁/c²/√1 - (v²/c²)]
= [(t₂ - t₁) - v/c²(x₂ - x₁)] / [√ 1 - v²/c²]
= {(1 × 10⁻³ s) - [(0.5 × 10⁸ m/s)/ (3× 10⁻⁸ m/s)²(10⁵m)}/ {√ 1 -[(0.5 ×10⁸ m/s)/3 ×10⁸m/s]²}
= (1 ms - 0.056ms) /√1 - 0.0278
= 0.944 / √ 1 - 0.0278
= 0.944/√0.9722
= 0.97ms/0.986
Answer = 0.983