Kvant Physics Problem 35

Verified: yes
Verdicts: PASS + PASS
Solve time: 12m03s
Source on kvant.digital

Problem

A satellite is flying at an altitude of 300 km. What stationary objects can be distinguished in a photograph taken from the satellite if the exposure time is 0.2 s?

Setup and Assumptions

The physical system consists of a satellite in low Earth orbit photographing the surface of the Earth. The satellite is flying at an altitude $h = 300~\text{km}$ above the Earth's surface. The exposure time of the photograph is $t_{\text{exp}} = 0.2~\text{s}$. The unknown is the minimum linear size $L_{\min}$ of stationary objects on the ground that can be distinguished in the photograph without significant motion blur caused by the satellite's motion. The satellite moves with orbital velocity $v_{\text{orb}}$ along a nearly circular orbit. The Earth is assumed to be spherical with radius $R_\oplus = 6371~\text{km}$. Atmospheric effects and optical aberrations are neglected. The camera is assumed to point toward the nadir. The smallest distinguishable object size is estimated by the displacement of the image of a ground point during the exposure.

Physical Principles

The orbital speed $v_{\text{orb}}$ of a satellite in a circular orbit at altitude $h$ is determined by the balance of gravitational and centripetal forces:

$$\frac{G M_\oplus m}{(R_\oplus + h)^2} = \frac{m v_{\text{orb}}^2}{R_\oplus + h},$$

where $G = 6.674 \times 10^{-11}\text{m}^3\text{kg}^{-1}\text{s}^{-2}$ and $M_\oplus = 5.972 \times 10^{24}\text{kg}$.

The relevant blur is not the distance traveled by the satellite itself. A ground point moves across the image according to the motion of the satellite's nadir point on the Earth's surface. If the satellite advances through a small orbital angle $d\theta$, then the satellite travels

$$(R_\oplus+h)d\theta,$$

while the corresponding displacement of the nadir point on the Earth's surface is

$$R_\oplus d\theta.$$

Hence the speed of the image across the ground is

$$v_g = \frac{R_\oplus}{R_\oplus+h},v_{\text{orb}}.$$

The motion blur during the exposure is then

$$s=v_g t_{\text{exp}}.$$

The minimum distinguishable object size is estimated by this blur length:

$$L_{\min}\approx s.$$

Derivation

The gravitational force acting on the satellite provides the necessary centripetal force for circular motion:

$$\frac{G M_\oplus m}{(R_\oplus + h)^2} = \frac{m v_{\text{orb}}^2}{R_\oplus + h}.$$

Cancelling $m$ and multiplying both sides by $(R_\oplus+h)$ gives

$$\frac{G M_\oplus}{R_\oplus+h} = v_{\text{orb}}^2.$$

Taking the square root yields

$$v_{\text{orb}} = \sqrt{\frac{G M_\oplus}{R_\oplus+h}}.$$

To relate satellite motion to image motion on the Earth's surface, consider a small angular displacement $d\theta$ during a time interval $dt$. The satellite displacement along its orbit is

$$ds_{\text{sat}}=(R_\oplus+h)d\theta,$$

so

$$v_{\text{orb}} = \frac{ds_{\text{sat}}}{dt} = (R_\oplus+h)\frac{d\theta}{dt}.$$

The corresponding displacement of the nadir point on the Earth's surface is

$$ds_g=R_\oplus d\theta,$$

hence

$$v_g = \frac{ds_g}{dt} = R_\oplus\frac{d\theta}{dt}.$$

Substituting

$$\frac{d\theta}{dt} = \frac{v_{\text{orb}}}{R_\oplus+h}$$

gives

$$v_g = \frac{R_\oplus}{R_\oplus+h},v_{\text{orb}}.$$

The blur length during the exposure is

$$s=v_g t_{\text{exp}} = t_{\text{exp}} \frac{R_\oplus}{R_\oplus+h} \sqrt{\frac{G M_\oplus}{R_\oplus+h}}.$$

The minimum distinguishable size is estimated by this blur:

$$L_{\min} \approx t_{\text{exp}} \frac{R_\oplus}{R_\oplus+h} \sqrt{\frac{G M_\oplus}{R_\oplus+h}}.$$

Result

Substituting

$$R_\oplus+h = 6371~\text{km}+300~\text{km} = 6671~\text{km} = 6.671\times10^6~\text{m}$$

gives

$$v_{\text{orb}} = \sqrt{\frac{6.674\times10^{-11}\cdot5.972\times10^{24}} {6.671\times10^6}} ~\text{m/s}.$$

The product in the numerator is

$$6.674\times10^{-11}\cdot5.972\times10^{24} = 3.986\times10^{14} ~\text{m}^3\text{s}^{-2}.$$

Dividing by $6.671\times10^6~\text{m}$ yields

$$\frac{3.986\times10^{14}} {6.671\times10^6} \approx 5.976\times10^7 ~\text{m}^2\text{s}^{-2}.$$

Taking the square root gives

$$v_{\text{orb}} \approx 7.73\times10^3~\text{m/s}.$$

The ground image speed is

$$v_g = \frac{6371}{6671} \cdot 7.73\times10^3 \approx 7.38\times10^3~\text{m/s}.$$

The blur during the $0.2~\text{s}$ exposure is

$$L_{\min} \approx s = v_g t_{\text{exp}} = (7.38\times10^3)(0.2) \text{m} \approx 1.48\times10^3\text{m}.$$

Thus the minimum size of stationary objects that can be distinguished is approximately

$$\boxed{L_{\min}\approx1.5~\text{km}}.$$

Sanity Checks

The expression

$$L_{\min} = t_{\text{exp}} \frac{R_\oplus}{R_\oplus+h} \sqrt{\frac{G M_\oplus}{R_\oplus+h}}$$

has dimensions of length because $\sqrt{GM_\oplus/(R_\oplus+h)}$ has units of velocity, the factor $R_\oplus/(R_\oplus+h)$ is dimensionless, and multiplication by time produces a length.

As $t_{\text{exp}}$ increases, the blur grows linearly, which matches the expected behavior of motion blur.

As $h$ increases, both the factor $R_\oplus/(R_\oplus+h)$ and the orbital speed decrease, so the blur decreases. This agrees with the fact that the ground image sweeps more slowly for a higher orbit.

The numerical value is about $1.5~\text{km}$, which is of the expected order of magnitude for a low Earth orbit satellite moving at several kilometers per second during a $0.2~\text{s}$ exposure.