Measuring an Event — Question 5

Section 37-2 · Physics pages 6060–6061

Question 5 A transponder delay biases naive radar range A stationary transponder waits δ = 2.0 ns after receiving a pulse before sending the reply. The source emits at zero and receives the reply at 22.0 ns. The delay is specified in the common rest frame. Use c = 3.00 × 108 m/s and Einstein-synchronized laboratory clocks. x ct cδ Spacetime sketch; the vertical coordinate is ct. Labels specify events and worldlines. (a) Find the actual range and the transponder reception time. (b) Find the error in range and reception time if the delay is ignored.
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Question 5 - Solution A transponder delay biases naive radar range x ct cδ (a) Method and calculation tR = 2L/c + δ, L = c(tR −δ)/2 = 3.00 m . tarrive = L/c = 10.0 ns . (b) Interpretation and further calculation Lnaive = ctR/2 = 3.30 m, ∆L = 0.30 m . tmidpoint = 11.0 ns, ∆t = 1.0 ns . Physical check The midpoint lies halfway through the waiting interval; it is neither the arrival nor the departure event.

Question 5

A transponder delay biases naive radar range

A stationary transponder waits δ=2.0ns\delta=2.0\,\mathrm{ns} after receiving a pulse before sending the reply. The source emits at zero and receives the reply at 22.0ns22.0\,\mathrm{ns}. The delay is specified in the common rest frame. Use c=3.00×108m/sc=3.00\times10^8\,\mathrm{m/s} and Einstein-synchronized laboratory clocks.

Diagram shown in the original SVG worksheet.

  1. Find the actual range and the transponder reception time.

  2. Find the error in range and reception time if the delay is ignored.

Question 5 - Solution

A transponder delay biases naive radar range

Diagram shown in the original SVG worksheet.

(a) Method and calculation

tR=2L/c+δ,L=c(tR−δ)/2=3.00m.t_R=2L/c+\delta,\qquad\boxed{L=c(t_R-\delta)/2=3.00\,\mathrm m}.tarrive=L/c=10.0ns.\boxed{t_{\mathrm{arrive}}=L/c=10.0\,\mathrm{ns}}.

(b) Interpretation and further calculation

Lnaive=ctR/2=3.30m,ΔL=0.30m.L_{\mathrm{naive}}=ct_R/2=3.30\,\mathrm m,\quad\boxed{\Delta L=0.30\,\mathrm m}.tmidpoint=11.0ns,Δt=1.0ns.t_{\mathrm{midpoint}}=11.0\,\mathrm{ns},\quad\boxed{\Delta t=1.0\,\mathrm{ns}}.

Physical check

The midpoint lies halfway through the waiting interval; it is neither the arrival nor the departure event.