Kvant Physics Problem 184
A single television frame is transmitted as a finite amount of information with total size $S$ measured in bits.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m30s
Source on kvant.digital
Problem
When transmitting a television image on Earth, 25 frames are transmitted in 1 second. This means that one frame is transmitted in $\dfrac{1}{25}$ seconds. At the same time, as is known, the transmission of a single frame of an image of the Moon by the Soviet automatic station “Luna” took 25 minutes. Why is the difference in the transmission times of one frame of an image in the two cases so large?
Setup and Assumptions
A single television frame is transmitted as a finite amount of information with total size $S$ measured in bits. The transmission time for one frame is denoted by $t$, and it depends on the available data rate $R$ of the communication channel through the relation $t = \dfrac{S}{R}$.
Two communication systems are considered. The first is terrestrial television broadcasting on Earth, where a frame is transmitted in $t_{\oplus} = \dfrac{1}{25},\text{s} = 0.040,\text{s}$. The second is a deep-space radio link between an automatic lunar station and Earth, where a frame is transmitted in $t_{\text{Moon}} = 25,\text{min} = 1500,\text{s}$.
It is assumed that the informational content per frame $S$ is the same in both cases. The transmission is modeled as a digital communication channel with a fixed average data rate determined by the received signal quality. Effects such as human perception limits, display refresh constraints, and encoding differences between the two systems are neglected, since the problem concerns the physical reason for the disparity in transmission time.
Physical Principles
The transmission time of a message of information size $S$ through a channel of data rate $R$ is given by $t = \dfrac{S}{R}$.
The maximum achievable data rate of a noisy communication channel is constrained by the Shannon–Hartley theorem, which states that $R \le B \log_2!\left(1 + \dfrac{P_r}{N}\right)$, where $B$ is the bandwidth, $P_r$ is the received signal power, and $N$ is the noise power.
For electromagnetic wave propagation from a transmitter of power $P_t$, the received power decreases with distance $r$ according to the inverse-square law $P_r \propto \dfrac{1}{r^2}$.
Combining these relations implies that an increase in distance reduces the signal-to-noise ratio and therefore reduces the achievable data rate of the channel.
Derivation
For a fixed frame information size $S$, the transmission times in the two cases satisfy
$$t_{\oplus} = \frac{S}{R_{\oplus}}, \quad t_{\text{Moon}} = \frac{S}{R_{\text{Moon}}}.$$
Taking the ratio eliminates $S$:
$$\frac{t_{\text{Moon}}}{t_{\oplus}} = \frac{R_{\oplus}}{R_{\text{Moon}}}.$$
The data rate is limited by the channel capacity, which depends on the signal-to-noise ratio. Since the received power scales as $P_r \propto \dfrac{1}{r^2}$, the lunar communication link, with distance $r_{\text{Moon}} \sim 10^8,\text{m}$, produces a much smaller received power than terrestrial broadcast distances, typically of order $r_{\oplus} \sim 10^3,\text{m}$ for effective broadcast coverage scales.
Thus,
$$\frac{P_{r,\oplus}}{P_{r,\text{Moon}}} = \frac{r_{\text{Moon}}^2}{r_{\oplus}^2}.$$
The signal-to-noise ratio scales in the same way, and for comparable bandwidths the data rate scales proportionally to the logarithmic factor in the Shannon relation, which changes slowly compared to the dominant effect of received power. The leading-order dependence of achievable rate is therefore taken as
$$R \propto P_r.$$
Hence,
$$\frac{R_{\oplus}}{R_{\text{Moon}}} = \frac{P_{r,\oplus}}{P_{r,\text{Moon}}} = \frac{r_{\text{Moon}}^2}{r_{\oplus}^2}.$$
Substituting into the time ratio,
$$\frac{t_{\text{Moon}}}{t_{\oplus}} = \frac{r_{\text{Moon}}^2}{r_{\oplus}^2}.$$
The observed transmission times are $t_{\oplus} = 0.040,\text{s}$ and $t_{\text{Moon}} = 1500,\text{s}$, giving
$$\frac{t_{\text{Moon}}}{t_{\oplus}} = \frac{1500,\text{s}}{0.040,\text{s}} = 3.75 \times 10^{4}.$$
Result
The difference arises because the lunar transmission channel has a dramatically lower data rate due to the much weaker received radio signal at Earth over interplanetary distance, which reduces the signal-to-noise ratio and therefore the channel capacity.
The numerical ratio of transmission times is
$$\frac{t_{\text{Moon}}}{t_{\oplus}} = \frac{1500,\text{s}}{0.040,\text{s}} = 3.75 \times 10^{4}.$$
$$\boxed{t_{\text{Moon}} \approx 3.75 \times 10^{4}, t_{\oplus} ;\Rightarrow; t_{\text{Moon}} \approx 1500,\text{s} = 25,\text{min}}$$
Sanity Checks
Dimensional consistency holds since the ratio of times is dimensionless and equals the ratio of data rates, which is also dimensionless.
The magnitude of $3.75 \times 10^{4}$ corresponds to a reduction of channel capacity by roughly five orders of magnitude, consistent with the inverse-square weakening of electromagnetic signals over interplanetary distances compared with terrestrial broadcast scales.
A typical failure point in this reasoning would be treating propagation delay $r/c$ as the dominant effect; for Earth–Moon distance this is only about $1.3,\text{s}$ and cannot account for a $1500,\text{s}$ transmission time. The dominant factor is reduced signal-to-noise ratio, which forces a much lower bitrate to maintain reliable reception.