Kvant Physics Problem 35

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Problem

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

Setup and Assumptions

A satellite moves in a circular low Earth orbit at altitude $h = 300,\text{km} = 3.0 \cdot 10^{5},\text{m}$ above Earth’s surface. The Earth’s radius is $R = 6.37 \cdot 10^{6},\text{m}$, so the orbital radius is $r = R + h$.

The exposure time of the camera is $t = 0.2,\text{s}$. During this time, the satellite moves relative to the ground, producing image smearing. The question asks for the characteristic size of stationary objects that can still be distinguished, which is determined by the ground distance covered by the satellite during exposure.

The satellite orbit is assumed circular, with motion governed only by Earth’s gravity. Atmospheric drag, attitude jitter, and optical resolution limits of the camera are neglected. The ground is treated as non-moving in the first approximation, with Earth’s rotation considered only as a secondary correction.

Physical Principles

For circular orbital motion, the gravitational force provides the centripetal force,

$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r},$

which gives the orbital speed

$v = \sqrt{\frac{GM}{r}}.$

The linear image smear on the ground during exposure is determined by uniform motion approximation,

$\Delta x = vt.$

An object can be distinguished as stationary if its characteristic size is comparable to or larger than the smear distance $\Delta x$.

Derivation

The orbital radius is

$r = R + h = 6.37 \cdot 10^{6},\text{m} + 3.0 \cdot 10^{5},\text{m} = 6.67 \cdot 10^{6},\text{m}.$

The standard gravitational parameter of Earth is $GM = 3.986 \cdot 10^{14},\text{m}^{3}\text{s}^{-2}$. The orbital speed follows from the balance of gravitational and centripetal acceleration,

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.986 \cdot 10^{14},\text{m}^{3}\text{s}^{-2}}{6.67 \cdot 10^{6},\text{m}}}.$

The ratio evaluates to

$\frac{3.986 \cdot 10^{14}}{6.67 \cdot 10^{6}} \approx 5.98 \cdot 10^{7},\text{m}^{2}\text{s}^{-2},$

so the orbital speed is

$v \approx \sqrt{5.98 \cdot 10^{7}},\text{m/s} \approx 7.73 \cdot 10^{3},\text{m/s}.$

The displacement of the ground projection during exposure is

$\Delta x = vt = (7.73 \cdot 10^{3},\text{m/s})(0.2,\text{s}) = 1.55 \cdot 10^{3},\text{m}.$

Result

The characteristic linear scale of stationary objects that can be distinguished is set by the motion blur distance,

$\Delta x = t\sqrt{\frac{GM}{R+h}}.$

Substitution gives

$\Delta x = (0.2,\text{s}) \cdot 7.73 \cdot 10^{3},\text{m/s} = 1.55 \cdot 10^{3},\text{m} = 1.55,\text{km}.$

The limiting distinguishable size of stationary objects is

$\boxed{1.6 \times 10^{3},\text{m} \approx 1.6,\text{km}}.$

Sanity Checks

The expression $\Delta x = vt$ has dimensions of length since $\text{m/s} \cdot \text{s} = \text{m}$, ensuring dimensional consistency.

The orbital speed scale of $7\text{–}8,\text{km/s}$ matches typical low Earth orbit values, so the displacement scale of order $10^{3},\text{m}$ over $0.2,\text{s}$ is consistent with expected satellite imaging blur.

Earth’s rotation produces a surface speed up to about $465,\text{m/s}$ at the equator, corresponding to an additional shift of $93,\text{m}$ during $0.2,\text{s}$, which is an order of magnitude smaller than the orbital contribution and does not change the kilometer-scale result. The dominant contribution arises from orbital motion, so the limiting resolution is governed by approximately $1.5,\text{km}$ ground displacement during exposure.