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

Valter edited Aug 21, 2026, 7:22 PM

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revision #20203 parent #20070 ← older

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			### Statement

		−	$2.6.35.$ Descending in a spiral from a circular orbit to the surface of ~~the~~ planet in the rarefied layers of the atmosphere, ~~the~~ satellite makes almost circular turns of decreasing radius. At the same time, its speed increases as if the force of atmospheric resistance pushes the satellite forward, in the direction of its flight! Explain qualitatively and quantitatively this paradoxical behavior of the satellite.
		+	$2.6.35^*.$ Descending in a spiral from a circular orbit to the surface of a planet in the rarefied layers of the atmosphere, a satellite makes almost circular turns of decreasing radius. At the same time, its speed increases as if the force of atmospheric resistance pushes the satellite forward, in the direction of its flight! Explain qualitatively and quantitatively this paradoxical behavior of the satellite.

			### Solution

		−	When a satellite with mass $m$ is in a circular orbit with radius $r$ around a planet with mass $M\gg m$, the gravitational force serves as the centripetal force
		+	<b>1. Qualitative explanation:</b>
		+	The atmospheric drag force does negative work, causing the total mechanical energy of the satellite to decrease. This loss of energy forces the satellite to drop deeper into the planet's gravitational potential well (the orbit radius decreases). As it approaches the planet, the gravitational force does positive work. The nature of the gravitational field is such that the work done by gravity during this descent is exactly twice the energy lost to atmospheric friction. As a result, the net kinetic energy of the satellite increases, and it accelerates.

		−	~~\[\frac~~{~~GMm~~}{r^2}=\frac{mv^2}{r}~~.\]~~
		+	<b>2. Quantitative explanation:</b>
		+	For a circular orbit of radius $r$, the velocity $v$ is determined from Newton's second law:
		+	$$G \frac{mM}{r^2} = m \frac{v^2}{r} \implies v^2 = \frac{GM}{r}$$
		+	Kinetic energy of the satellite: $E_k = \frac{mv^2}{2} = \frac{GMm}{2r}$.
		+	Potential energy of the satellite in the gravitational field: $U = -\frac{GMm}{r}$.
		+	Total mechanical energy:
		+	$$E = E_k + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$$
		+	Note the fundamental relationship: $E = -E_k$.

		−	~~In this orbit, the gravitational potential energy is~~
		+	The change in total energy $dE$ due to the work of the drag force $F_{\text{drag}}$ over a small path element $ds$ is:
		+	$$dE = -F_{\text{drag}} ds$$
		+	Since $E = -E_k$, by differentiating we get $dE_k = -dE$. Substituting this into the work equation gives:
		+	$$dE_k = F_{\text{drag}} ds$$
		+	Let's express the increment of kinetic energy via the differential $dE_k = d\left(\frac{mv^2}{2}\right) = m v dv$:
		+	$$m v dv = F_{\text{drag}} ds$$
		+	Dividing both sides by $dt$ (considering that $v = \frac{ds}{dt}$), we find the effective tangential acceleration of the satellite:
		+	$$m \frac{dv}{dt} = F_{\text{drag}} \implies a_{\tau} = \frac{F_{\text{drag}}}{m}$$

		−	\[-\frac{GMm}{r}=-2\cdot\frac{1}{2}mv^2;\]
		−
		−	that is, the gravitational potential energy is negative but its magnitude is twice the kinetic energy. Consequently, if the total energy of the satellite changes by $-\Delta E$, then the gravitational potential energy changes by $-2\cdot\Delta E$ and the kinetic energy changes by $\Delta E$. Therefore, the speed of the satellite increases even though its total energy decreases.
		−
		−	In terms of forces, the atmospheric resistance does decrease the tangential velocity of the satellite. However, the decrese in orbital radius means that the gravitational force increases the inward radial velocity, resulting in an overall increase in the total speed of the satellite.
		−
			#### Answer
		−
		−	See the solution.
		+	From the relationship between total and kinetic energies ($E = -E_k$), it follows that $dE_k = -dE$. Since the drag force decreases the total energy ($dE = -F_{\text{drag}} ds$), the kinetic energy increases by $dE_k = F_{\text{drag}} ds$. Mathematically, this looks exactly as if the drag force $F_{\text{drag}}$ is applied in the direction of the velocity vector and accelerates the satellite.

	@@ -1,21 +1,29 @@
	### Statement		### Statement

	$2.6.35.$ Descending in a spiral from a circular orbit to the surface of ~~the~~ planet in the rarefied layers of the atmosphere, ~~the~~ satellite makes almost circular turns of decreasing radius. At the same time, its speed increases as if the force of atmospheric resistance pushes the satellite forward, in the direction of its flight! Explain qualitatively and quantitatively this paradoxical behavior of the satellite.		$2.6.35^*.$ Descending in a spiral from a circular orbit to the surface of a planet in the rarefied layers of the atmosphere, a satellite makes almost circular turns of decreasing radius. At the same time, its speed increases as if the force of atmospheric resistance pushes the satellite forward, in the direction of its flight! Explain qualitatively and quantitatively this paradoxical behavior of the satellite.

	### Solution		### Solution

	When a satellite with mass $m$ is in a circular orbit with radius $r$ around a planet with mass $M\gg m$, the gravitational force serves as the centripetal force		<b>1. Qualitative explanation:</b>
			The atmospheric drag force does negative work, causing the total mechanical energy of the satellite to decrease. This loss of energy forces the satellite to drop deeper into the planet's gravitational potential well (the orbit radius decreases). As it approaches the planet, the gravitational force does positive work. The nature of the gravitational field is such that the work done by gravity during this descent is exactly twice the energy lost to atmospheric friction. As a result, the net kinetic energy of the satellite increases, and it accelerates.

	~~\[\frac~~{~~GMm~~}{r^2}=\frac{mv^2}{r}~~.\]~~		<b>2. Quantitative explanation:</b>
			For a circular orbit of radius $r$, the velocity $v$ is determined from Newton's second law:
			$$G \frac{mM}{r^2} = m \frac{v^2}{r} \implies v^2 = \frac{GM}{r}$$
			Kinetic energy of the satellite: $E_k = \frac{mv^2}{2} = \frac{GMm}{2r}$.
			Potential energy of the satellite in the gravitational field: $U = -\frac{GMm}{r}$.
			Total mechanical energy:
			$$E = E_k + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$$
			Note the fundamental relationship: $E = -E_k$.

	~~In this orbit, the gravitational potential energy is~~		The change in total energy $dE$ due to the work of the drag force $F_{\text{drag}}$ over a small path element $ds$ is:
			$$dE = -F_{\text{drag}} ds$$
			Since $E = -E_k$, by differentiating we get $dE_k = -dE$. Substituting this into the work equation gives:
			$$dE_k = F_{\text{drag}} ds$$
			Let's express the increment of kinetic energy via the differential $dE_k = d\left(\frac{mv^2}{2}\right) = m v dv$:
			$$m v dv = F_{\text{drag}} ds$$
			Dividing both sides by $dt$ (considering that $v = \frac{ds}{dt}$), we find the effective tangential acceleration of the satellite:
			$$m \frac{dv}{dt} = F_{\text{drag}} \implies a_{\tau} = \frac{F_{\text{drag}}}{m}$$

	\[-\frac{GMm}{r}=-2\cdot\frac{1}{2}mv^2;\]

	that is, the gravitational potential energy is negative but its magnitude is twice the kinetic energy. Consequently, if the total energy of the satellite changes by $-\Delta E$, then the gravitational potential energy changes by $-2\cdot\Delta E$ and the kinetic energy changes by $\Delta E$. Therefore, the speed of the satellite increases even though its total energy decreases.

	In terms of forces, the atmospheric resistance does decrease the tangential velocity of the satellite. However, the decrese in orbital radius means that the gravitational force increases the inward radial velocity, resulting in an overall increase in the total speed of the satellite.

	#### Answer		#### Answer
			From the relationship between total and kinetic energies ($E = -E_k$), it follows that $dE_k = -dE$. Since the drag force decreases the total energy ($dE = -F_{\text{drag}} ds$), the kinetic energy increases by $dE_k = F_{\text{drag}} ds$. Mathematically, this looks exactly as if the drag force $F_{\text{drag}}$ is applied in the direction of the velocity vector and accelerates the satellite.
	See the solution.