Edits to “Statement”, “Solution”, “Answer”

jzmicer edited Jun 16, 2026, 8:28 PM

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revision #18946 parent #18928 ← older

en/12.2.14.md +39 −7

		@@ -1,15 +1,47 @@
			### Statement

		−	$12.2.14.$ [Insert the problem statement]
		+	$12.2.14.$ Estimate the minimum size of an object on the Earth’s surface that can be photographed from a satellite flying at an altitude of $200 \ km$, as well as the minimum size of objects on the Moon and on Mars that can be photographed from near-Earth orbit. The resolution of the film does not limit the clarity of the image.

			### Solution

		−	~~Estimate the minimum size of an object on the Earth’s surface that can be~~
		−	photographed from a satellite flying at an altitude of 200 km, as well as the
		−	minimum size of objects on the Moon and on Mars that can be photographed
		−	from near-Earth orbit. The resolution of the film does not limit the clarity of
		−	the image.
		+	When light from an object passes through the entrance aperture of the camera, a diffraction pattern, not a point, is formed on the sensor.

		+	We are not photographing a single isolated point (it can always be seen as a small spot, even if it is blurred).
		+	We need to distinguish two close details (for example, two craters or two closely spaced objects).
		+	If two objects are too close, their diffraction spots overlap, and we see one blurred spot instead of two.
		+
		+	Two point sources are considered resolved if the central maximum of one source falls exactly on the first minimum of the diffraction pattern of the other. The same condition is necessary to see the edges of our object separately. Therefore, the object must be seen from the shooting point at an angle greater than some $\theta_{min}$, where $\theta_{min}$ is the angle corresponding to the first diffraction minimum.
		+
		+	We take the distance from the Earth to the Moon as $384\,467$ km, and to Mars as $228.4$ million km.
		+
		+	We also assume that we managed to buy a camera with a slit‑shaped lens: this will greatly simplify the mathematics. Since we are only required to make an estimate, this is acceptable. Let the slit width $d$ be $d = 15$ cm. To obtain an upper estimate, we take the wavelength corresponding to the upper limit of the visible range, $\lambda = 750$ nm.
		+
		+	After these questionable simplifications, we obtain the system described in problem [12.2.4](https://savchenkosolutions.com/en/12.2.4), from which we can directly take the formula for $\sin\theta_{min}$, keeping in mind that for huge distances in space $\sin\theta_{min}\approx \theta_{min}$ holds:
		+
		+	$$
		+	\theta_{min} = \frac{\lambda}{d}.
		+	$$
		+
		+	Then the size of an object located at a distance $r$ is:
		+	$$
		+	l \approx r\theta_{min} = r\frac{\lambda}{d}.
		+	$$
		+
		+	And the calculations:
		+	$$
		+	l_{\text{satellite – Earth}} \approx 1\ \text{m}
		+	$$
		+	$$
		+	l_{\text{Earth – Moon}} \approx 2\ \text{km}
		+	$$
		+	$$
		+	l_{\text{Earth – Mars}} \approx 1142\ \text{km}
		+	$$
		+
		+	Savchenko's answer is slightly different – the estimate strongly depends on the chosen values (especially the distance to Mars, which varies considerably during the year); in order of magnitude, everything agrees more or less. The framed text is the author's answer.
		+
			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$$
		+	\boxed{l \approx 1 \ m, 0.5 \ km, 150 \ km.}
		+	$$

	@@ -1,15 +1,47 @@
	### Statement		### Statement

	$12.2.14.$ [Insert the problem statement]		$12.2.14.$ Estimate the minimum size of an object on the Earth’s surface that can be photographed from a satellite flying at an altitude of $200 \ km$, as well as the minimum size of objects on the Moon and on Mars that can be photographed from near-Earth orbit. The resolution of the film does not limit the clarity of the image.

	### Solution		### Solution

	~~Estimate the minimum size of an object on the Earth’s surface that can be~~		When light from an object passes through the entrance aperture of the camera, a diffraction pattern, not a point, is formed on the sensor.
	photographed from a satellite flying at an altitude of 200 km, as well as the
	minimum size of objects on the Moon and on Mars that can be photographed
	from near-Earth orbit. The resolution of the film does not limit the clarity of
	the image.

			We are not photographing a single isolated point (it can always be seen as a small spot, even if it is blurred).
			We need to distinguish two close details (for example, two craters or two closely spaced objects).
			If two objects are too close, their diffraction spots overlap, and we see one blurred spot instead of two.

			Two point sources are considered resolved if the central maximum of one source falls exactly on the first minimum of the diffraction pattern of the other. The same condition is necessary to see the edges of our object separately. Therefore, the object must be seen from the shooting point at an angle greater than some $\theta_{min}$, where $\theta_{min}$ is the angle corresponding to the first diffraction minimum.

			We take the distance from the Earth to the Moon as $384\,467$ km, and to Mars as $228.4$ million km.

			We also assume that we managed to buy a camera with a slit‑shaped lens: this will greatly simplify the mathematics. Since we are only required to make an estimate, this is acceptable. Let the slit width $d$ be $d = 15$ cm. To obtain an upper estimate, we take the wavelength corresponding to the upper limit of the visible range, $\lambda = 750$ nm.

			After these questionable simplifications, we obtain the system described in problem [12.2.4](https://savchenkosolutions.com/en/12.2.4), from which we can directly take the formula for $\sin\theta_{min}$, keeping in mind that for huge distances in space $\sin\theta_{min}\approx \theta_{min}$ holds:

			$$
			\theta_{min} = \frac{\lambda}{d}.
			$$

			Then the size of an object located at a distance $r$ is:
			$$
			l \approx r\theta_{min} = r\frac{\lambda}{d}.
			$$

			And the calculations:
			$$
			l_{\text{satellite – Earth}} \approx 1\ \text{m}
			$$
			$$
			l_{\text{Earth – Moon}} \approx 2\ \text{km}
			$$
			$$
			l_{\text{Earth – Mars}} \approx 1142\ \text{km}
			$$

			Savchenko's answer is slightly different – the estimate strongly depends on the chosen values (especially the distance to Mars, which varies considerably during the year); in order of magnitude, everything agrees more or less. The framed text is the author's answer.

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$$
			\boxed{l \approx 1 \ m, 0.5 \ km, 150 \ km.}
			$$