A spacecraft with a proper length of 400 m passes by an observer on the Earth. According to this observer, it takes 0.800 µs for the spacecraft to pass a fixed point. Determine the speed of the spacecraft as measured by the Earth-based observer. 0.857c
The correct answer and explanation is:
To determine the speed of the spacecraft as observed from Earth, we use the concept of length contraction and relative velocity from special relativity. However, in this case, we are given the proper length (length measured in the rest frame of the spacecraft) and the time interval it takes to pass a fixed point on Earth. From the Earth frame, the spacecraft is moving, so we can directly use the classical formula for speed:
Given:
- Proper length of the spacecraft (length in spacecraft’s frame), L0=400 mL_0 = 400 \, \text{m}
- Time taken to pass a fixed point on Earth, t=0.800 μs=0.800×10−6 st = 0.800 \, \mu s = 0.800 \times 10^{-6} \, \text{s}
Step 1: Find the contracted length LL
We are observing from Earth, so the moving spacecraft appears contracted. However, in this problem, we don’t need to apply length contraction directly, because we are given the proper length and time it takes to pass a fixed point (meaning only the front and back of the spacecraft pass the point in that time), so we can treat L0L_0 as the distance traveled in that time from Earth’s point of view.
Step 2: Use speed formula
v=distancetime=L0tv = \frac{\text{distance}}{\text{time}} = \frac{L_0}{t} v=400 m0.800×10−6 s=5.0×108 m/sv = \frac{400 \, \text{m}}{0.800 \times 10^{-6} \, \text{s}} = 5.0 \times 10^8 \, \text{m/s}
Step 3: Express as a fraction of the speed of light
Speed of light, c=3.0×108 m/sc = 3.0 \times 10^8 \, \text{m/s} vc=5.0×1083.0×108=1.67\frac{v}{c} = \frac{5.0 \times 10^8}{3.0 \times 10^8} = 1.67
But this result is not physically possible, since nothing can exceed the speed of light. This implies we misinterpreted the given length.
Important Clarification:
The proper length of 400 m is measured in the spacecraft’s rest frame. To an observer on Earth, the spacecraft is contracted in length due to its relativistic speed. The time given (0.800 µs) is the time measured in the Earth frame, which is how long the contracted spacecraft takes to pass the point.
Let LL be the contracted length as seen from Earth. Then: L=vtL = v t
But the contracted length is also given by: L=L0γ,where γ=11−(vc)2L = \frac{L_0}{\gamma}, \quad \text{where } \gamma = \frac{1}{\sqrt{1 – \left( \frac{v}{c} \right)^2 }}
So we have: vt=L0γ⇒v=L0γtv t = \frac{L_0}{\gamma} \Rightarrow v = \frac{L_0}{\gamma t}
Multiply both sides by γ\gamma: vγ=L0tv \gamma = \frac{L_0}{t}
Now substitute γ=11−(v/c)2\gamma = \frac{1}{\sqrt{1 – (v/c)^2}}: v⋅11−(v/c)2=4000.800×10−6=5.0×108v \cdot \frac{1}{\sqrt{1 – (v/c)^2}} = \frac{400}{0.800 \times 10^{-6}} = 5.0 \times 10^8
Let’s solve for vv numerically. Try:
Trial: v=0.857cv = 0.857c
Then: v=0.857×3.0×108=2.571×108 m/sv = 0.857 \times 3.0 \times 10^8 = 2.571 \times 10^8 \, \text{m/s} γ=11−(0.857)2=11−0.734=10.266≈1.937\gamma = \frac{1}{\sqrt{1 – (0.857)^2}} = \frac{1}{\sqrt{1 – 0.734}} = \frac{1}{\sqrt{0.266}} \approx 1.937
Now compute: vγ=2.571×108×1.937≈4.98×108≈5.0×108v \gamma = 2.571 \times 10^8 \times 1.937 \approx 4.98 \times 10^8 \approx 5.0 \times 10^8
✅ This matches the required value.
✅ Final Answer:
v=0.857c\boxed{v = 0.857c}
Explanation (300 words):
In this problem, we’re analyzing how fast a spacecraft must be moving for it to pass a fixed point on Earth in a very short time, given that its proper length is 400 meters. The term proper length refers to the length measured in the frame in which the object (spacecraft) is at rest — in this case, 400 meters.
However, observers on Earth see the spacecraft moving at a high speed, which means they perceive it to be length-contracted due to the effects of special relativity. At relativistic speeds (close to the speed of light), lengths contract along the direction of motion when measured from a different inertial frame.
The observer on Earth sees the spacecraft pass by in just 0.800 microseconds. To find the speed, we must relate the proper length to the contracted length observed from Earth. Using the Lorentz transformation, we know that: L=L0γ,andL=vtL = \frac{L_0}{\gamma}, \quad \text{and} \quad L = vt
Combining the two gives: vt=L0γ⇒vγ=L0tvt = \frac{L_0}{\gamma} \Rightarrow v\gamma = \frac{L_0}{t}
By solving this numerically and testing possible values, we find that when v=0.857cv = 0.857c, the equation is satisfied. This confirms the spacecraft is moving at 85.7% of the speed of light. This scenario illustrates how relativistic effects like time dilation and length contraction become significant at high velocities, which is a fundamental prediction of Einstein’s theory of special relativity.