Kvant Physics Problem 181

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Problem

The problem was mistakenly published under number 183 in the journal.

A satellite moves around the Earth in a nearly circular orbit with speed $v$. The change in its orbit is due to the fact that the satellite is acted upon by a friction force from microparticles $F = A \cdot v^\alpha$, where $A$ and $\alpha$ are constants. Find $\alpha$, if it is known that the radius of the satellite’s orbit changes very slowly.

Setup and Assumptions

A satellite of mass $m$ moves around the Earth of mass $M$ in a nearly circular orbit of radius $r$ with orbital speed $v$. The gravitational parameter is $\mu = GM$ with units $\mathrm{m^3,s^{-2}}$. A resistive force due to collisions with microparticles acts opposite to the velocity and has magnitude $F = A v^\alpha$, where $A$ is a constant with units $\mathrm{kg,m^{1-\alpha},s^{\alpha-2}}$ and $\alpha$ is dimensionless.

The orbit is assumed to remain circular at each instant, while the radius changes slowly in time so that at every moment the condition of circular motion is satisfied with negligible radial acceleration beyond gravity and centrifugal balance. The resistive force is taken to be purely tangential, so it does not directly modify the radial force balance. Effects such as Earth’s rotation, radiation pressure, and non-uniformity of the medium are neglected.

The unknown is the exponent $\alpha$.

Physical Principles

Newton’s second law in the tangential direction gives the evolution of orbital energy through the work of non-conservative forces, so that

$\frac{dE}{dt} = \mathbf{F}\cdot \mathbf{v}.$

For a circular Keplerian orbit, the speed and radius satisfy the balance of centripetal and gravitational accelerations,

$\frac{v^2}{r} = \frac{\mu}{r^2},$

so that

$v^2 = \frac{\mu}{r}.$

The total mechanical energy and angular momentum of a circular orbit are

$E = -\frac{\mu m}{2r}, \qquad L = m v r.$

The torque of the resistive force determines the angular momentum loss,

$\frac{dL}{dt} = -F r.$

Consistency of a slowly varying circular orbit requires that the time derivatives obtained from orbital relations $E(r)$ and $L(r)$ coincide with those obtained from the dissipative force laws.

Derivation

The circular-orbit condition gives the velocity as a function of radius,

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

The orbital energy expressed through $r$ is

$E(r) = -\frac{\mu m}{2r},$

so its derivative is

$\frac{dE}{dr} = \frac{\mu m}{2r^2}.$

The angular momentum is

$L(r) = m v r = m r \sqrt{\frac{\mu}{r}} = m \sqrt{\mu r},$

and differentiation yields

$\frac{dL}{dr} = \frac{m}{2}\sqrt{\frac{\mu}{r}} = \frac{L}{2r}.$

The dissipative force magnitude is $F = A v^\alpha$, so the energy loss rate due to work is

$\frac{dE}{dt} = -Fv = -A v^{\alpha+1}.$

Substituting $v = (\mu/r)^{1/2}$ gives

$\frac{dE}{dt} = -A \left(\frac{\mu}{r}\right)^{\frac{\alpha+1}{2}} = -A \mu^{\frac{\alpha+1}{2}} r^{-\frac{\alpha+1}{2}}.$

The torque equation gives

$\frac{dL}{dt} = -F r = -A v^\alpha r.$

Substitution of $v$ yields

$\frac{dL}{dt} = -A \left(\frac{\mu}{r}\right)^{\frac{\alpha}{2}} r = -A \mu^{\frac{\alpha}{2}} r^{1-\frac{\alpha}{2}}.$

Using $L(r)$, the time derivative of angular momentum from orbital geometry is

$\frac{dL}{dt} = \frac{dL}{dr}\frac{dr}{dt} = \frac{L}{2r}\frac{dr}{dt}.$

Equating this with the torque expression gives

$\frac{L}{2r}\frac{dr}{dt} = -A \mu^{\frac{\alpha}{2}} r^{1-\frac{\alpha}{2}}.$

Substituting $L = m\sqrt{\mu r}$ leads to

$\frac{m\sqrt{\mu r}}{2r}\frac{dr}{dt} = -A \mu^{\frac{\alpha}{2}} r^{1-\frac{\alpha}{2}}.$

Solving for $\frac{dr}{dt}$ gives

$\frac{dr}{dt} = -\frac{2A}{m}\mu^{\frac{\alpha-1}{2}} r^{\frac{3-\alpha}{2}}.$

The energy relation gives

$\frac{dE}{dt} = \frac{dE}{dr}\frac{dr}{dt} = \frac{\mu m}{2r^2}\frac{dr}{dt}.$

Substituting $\frac{dr}{dt}$ yields

$\frac{dE}{dt} = \frac{\mu m}{2r^2}\left(-\frac{2A}{m}\mu^{\frac{\alpha-1}{2}} r^{\frac{3-\alpha}{2}}\right).$

After cancellation of $m$ and $2$, this becomes

$\frac{dE}{dt} = -A \mu^{\frac{\alpha+1}{2}} r^{-\frac{\alpha+1}{2}}.$

This expression coincides identically with the direct work formula obtained from the force,

$\frac{dE}{dt} = -A \mu^{\frac{\alpha+1}{2}} r^{-\frac{\alpha+1}{2}}.$

Thus the circular-orbit evolution is dynamically consistent for any exponent $\alpha$, and the value of $\alpha$ must be fixed by the physical mechanism producing the drag.

For a satellite moving through a dilute medium of microparticles at rest in the Earth frame, the momentum transfer rate is proportional to the incident particle flux, which scales as $v$, while the momentum change per collision also scales as $v$. The resulting resistive force scales as $F \propto v^2$, giving $\alpha = 2$.

Result

$\alpha = 2.$

Substitution requires no numerical parameters since $\alpha$ is dimensionless.

$$\boxed{\alpha = 2}$$

Sanity Checks

Dimensional consistency requires $A v^\alpha$ to have units of force $\mathrm{N} = \mathrm{kg,m,s^{-2}}$, which fixes the units of $A$ as $\mathrm{kg,m^{1-\alpha},s^{\alpha-2}}$, consistent for $\alpha=2$.

For $\alpha=2$, the drag force becomes quadratic in velocity, matching the standard regime of high-speed motion through a dilute gas or particle medium where momentum flux scales with $v^2$.

The derived evolution law $\frac{dr}{dt} \propto r^{(3-\alpha)/2}$ becomes $\frac{dr}{dt} \propto r^{1/2}$ for $\alpha=2$, producing a gradual inward spiral consistent with the assumption of slow orbital radius variation.

Any sign error would most easily enter in the torque relation $\frac{dL}{dt} = -Fr$, since reversing the direction of the tangential force would invert the inferred decay of $L$ and $r$ while leaving the energy balance formally unchanged.