Kvant Physics Problem 184
The physical system is a television image transmission channel.
Verified: yes
Verdicts: PASS + PASS
Solve time: 9m13s
Source on kvant.digital
Problem
When transmitting a television image on Earth, 25 frames are transmitted per second. This means that one frame is transmitted in $\dfrac{1}{25}$ seconds. At the same time, as is well known, the transmission of a single frame of an image of the Moon by the Soviet automatic station Luna took 25 minutes. Why is there such a large difference in the transmission times for one frame of an image in these two cases?
Setup and Assumptions
The physical system is a television image transmission channel. The quantity to be explained is the time required to transmit one complete frame of an image.
For ordinary terrestrial television, the frame transmission time is
$$t_{\text{TV}}=\frac{1}{25},\text{s}=0.04,\text{s}.$$
For the transmission of a lunar image by the automatic station Luna, the frame transmission time was
$$t_{\text{Luna}}=25,\text{min}=1500,\text{s}.$$
The ratio of these times is
$$\frac{t_{\text{Luna}}}{t_{\text{TV}}} = \frac{1500}{0.04} = 3.75\times10^{4}.$$
The transmission path itself is not the determining factor. Electromagnetic waves travel with speed close to
$$c \approx 3\times10^{8},\text{m/s}.$$
The mean Earth-Moon distance is approximately
$$R \approx 3.8\times10^{8},\text{m}.$$
The analysis assumes that the principal limitation is the amount of information that must be sent per second. Effects associated with propagation losses, equipment imperfections, and signal processing details are neglected unless they influence the information rate.
Physical Principles
The time required to transmit an image is determined by the amount of information contained in the image and by the communication channel's information rate.
If an image contains an information volume $I$ and the communication channel transmits information at rate $B$, then
$$t=\frac{I}{B}.$$
For electromagnetic signals, the propagation time from source to receiver is
$$t_{\text{prop}}=\frac{R}{c}.$$
A communication channel with a smaller available bandwidth can transmit less information per second, increasing the time required to send a frame.
Derivation
First, evaluate whether the large difference in frame transmission times can be explained simply by the signal travel time.
For transmission from the Moon,
$$t_{\text{prop}} = \frac{R}{c} = \frac{3.8\times10^{8},\text{m}} {3\times10^{8},\text{m/s}} \approx 1.3,\text{s}.$$
The one-way propagation delay is only about one second.
The observed frame transmission time is
$$1500,\text{s},$$
which exceeds the propagation delay by a factor
$$\frac{1500}{1.3} \approx 1.2\times10^{3}.$$
The enormous transmission time cannot be caused by the finite speed of radio waves.
The explanation must be sought in the information rate. A terrestrial television station operates with a very wide transmission bandwidth and high signal power. Such a channel can carry the information of a complete frame in
$$0.04,\text{s}.$$
A spacecraft near the Moon has severe limitations. Its transmitter power is small, its antenna is limited in size, and the received signal on Earth is extremely weak after traveling hundreds of thousands of kilometers. To ensure reliable reception, the image information must be transmitted much more slowly, using a far smaller effective information rate.
Let $I$ denote the information contained in one frame. Then
$$t_{\text{TV}}=\frac{I}{B_{\text{TV}}}, \qquad t_{\text{Luna}}=\frac{I}{B_{\text{Luna}}}.$$
Dividing these relations gives
$$\frac{t_{\text{Luna}}}{t_{\text{TV}}} = \frac{B_{\text{TV}}}{B_{\text{Luna}}}.$$
Using the given times,
$$\frac{B_{\text{TV}}}{B_{\text{Luna}}} = 3.75\times10^{4}.$$
Thus the lunar image channel transmitted information roughly tens of thousands of times more slowly than an ordinary television channel.
Result
The transmission time is governed primarily by the information rate of the communication channel,
$$t=\frac{I}{B}.$$
The Moon-Earth propagation delay is only
$$t_{\text{prop}} \approx \frac{3.8\times10^{8},\text{m}} {3\times10^{8},\text{m/s}} \approx 1.3,\text{s},$$
which is negligible compared with the observed
$$25,\text{min}=1500,\text{s}.$$
The ratio of transmission times is
$$\frac{1500,\text{s}}{0.04,\text{s}} = 3.75\times10^{4}.$$
Hence the lunar communication link had an information transmission rate approximately
$$3.75\times10^{4}$$
times smaller than that of ordinary television.
$$\boxed{\text{The large difference is not due to the distance to the Moon; it is due to the much lower information rate of the lunar radio channel.}}$$
Sanity Checks
The dimensions of the relation
$$t=\frac{I}{B}$$
are correct because information divided by information per unit time gives time.
If the communication rate $B$ were increased, the transmission time would decrease in inverse proportion. This agrees with the operation of ordinary television systems.
If distance alone were responsible, the relevant time scale would be the propagation delay of about $1,\text{s}$, not $1500,\text{s}$. The discrepancy of more than three orders of magnitude shows that propagation time is not the controlling factor.
The most common mistake is to attribute the 25-minute interval to the finite speed of light. Computing
$$R/c \approx 1.3,\text{s}$$
immediately reveals that this interpretation is incorrect.