91Ó°ÊÓ

The period between the application of the radiofrequency excitation pulse and the peak of the signal created in the coil is referred to as the echo time.

(a)

The echo time is the time it takes for light to travel from the source to the destination ${\mathrm{t}}_{\mathrm{echo}}=1000 \mathrm{s}.$We want to know how long it takes for light to go to one item (but not return), thus$\mathrm{t} =\frac{{\mathrm{t}}_{\mathrm{echo}}}{2}$ is the answer. As we know the speed of light$\mathrm{c}=3×{10}^8{\mathrm{ms}}^{-1}$ , we can calculate the distance to the planet d .

$$\begin{gathered} \mathrm{d}=\mathrm{c}×\mathrm{t} \\ =(3×{10}^8\mathrm{m}/\mathrm{s})×(500\mathrm{s}) \\ =1.5×{10}^{11}\mathrm{m} \end{gathered}$$

Therefore, the distance is$1.5×{10}^{11}\mathrm{m}$ .

(b)

We can compute a one-way journey time for light using the same method as in (a) and multiply it by 2 to get the round-trip time. We may use c it to determine the echo time since the distance to the automobile is d = 75.0m .

$$\begin{gathered} {\mathrm{t}}_{\mathrm{echo}}=2×\mathrm{t} \\ =2×\frac{\mathrm{d}}{\mathrm{c}} \\ =2×\frac{(75.0\mathrm{m})}{(3×{10}^8\mathrm{m}/\mathrm{s})} \\ =5×{10}^{-7}\mathrm{s} \end{gathered}$$

Therefore, the echo time is $5×{10}^{-7}\mathrm{s}$ .

(c)

To compute the distance with$\mathrm{Δ»å}=10.0 \mathrm{m}$accuracy, one must calculate how the echo time varies when the distance is increased by 10.0 m.

$$\begin{gathered} ∆{\mathrm{t}}_{\mathrm{echo}}=2×\frac{∆\mathrm{d}}{\mathrm{c}} \\ =2×\frac{(10.0\mathrm{m})}{\left((3×{10}^8\mathrm{m}/\mathrm{s})\right)} \\ =6.7×{10}^{-8}\mathrm{s}\left(\frac{1\mathrm{ns}}{{10}^{-9}\mathrm{s}}\right) \\ =67\mathrm{ns} \end{gathered}$$

Therefore, the echo time is $67\mathrm{ns}$ .