If a clock in an airplane is found to slow down by 5 parts in 1013, (i.e., ∆t/∆to = 1.0 + 5.0 × 10. −13), at what speed is the airplane traveling? (Hint: You might need to use an expansion for y)
I am not sure which equation to use can someone please help me.
L=L0(square root 1-v'2/c'2)
&=1/square root 1-v'2/c'2
use the second equation.
The first equation lhas to do with length contraction. Use the second one
Set 1 + 5.0*10^-13 = 1/sqrt[1 - (v/c)^2]
Then solve for v, or v/c
To determine the velocity of the airplane, we can use the time dilation equation. The equation relates the observed time interval (∆t) to the proper time interval (∆t0) when an object is in motion relative to an observer:
∆t = ∆t0 / √(1 - (v'/c')^2)
Here, v' represents the velocity of the airplane, and c' is the speed of light.
Given that the clock in the airplane slows down by 5 parts in 10^13, we can rewrite the time dilation equation as:
1.0 + 5.0 × 10^-13 = 1 / √(1 - (v'/c')^2)
Now, we need to use a small-angle approximation for square roots. It states that if x is a small number, then sqrt(1 + x) ≈ 1 + 0.5x.
Using this approximation, we can rewrite the equation as:
1.0 + 5.0 × 10^-13 ≈ 1 + 0.5 × (v'/c')^2
Since the velocity of the airplane is much smaller than the speed of light (v' << c'), we can neglect the terms with higher powers of (v'/c').
Therefore, we have:
5.0 × 10^-13 ≈ 0.5 × (v'/c')^2
Simplifying this equation, we find:
(v'/c')^2 ≈ (5.0 × 10^-13) / 0.5
(v'/c')^2 ≈ 1.0 × 10^-12
Taking the square root of both sides, we get:
v'/c' ≈ √(1.0 × 10^-12)
v'/c' ≈ 1.0 × 10^-6
Finally, rearranging the equation, we find the velocity of the airplane:
v' ≈ c' × (1.0 × 10^-6)
v' ≈ (3.0 × 10^8 m/s) × (1.0 × 10^-6)
v' ≈ 300 m/s
Therefore, the airplane is traveling at approximately 300 meters per second.