Kvant Physics Problem 181

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Verdicts: UNKNOWN + UNKNOWN
Solve time: 27m40s
Source on kvant.digital

Problem

In the journal it was mistakenly published under number 183.

A satellite moves around the Earth in an almost 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 $\alpha$, given that the orbital radius of the satellite 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 interaction obeys Newton's law with gravitational parameter $GM$. In addition, the satellite experiences a tangential drag force from microparticles, given by $F = A v^\alpha$, acting opposite to its velocity, where $A$ and $\alpha$ are constants. The orbit is assumed to remain circular at each instant, with the radius changing slowly compared to the orbital period. The mass of the satellite remains constant, and other non-gravitational forces are neglected.

Physical Principles

For a circular orbit, the centripetal acceleration is balanced by gravitational attraction, giving

$\frac{m v^2}{r} = \frac{GM m}{r^2},$

which implies

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

The total mechanical energy of a circular orbit is

$E = -\frac{GM m}{2 r},$

and the orbital angular momentum is

$L = m v r = m \sqrt{GM r}.$

The tangential drag force performs work on the satellite, producing a rate of change of energy

$\frac{dE}{dt} = \mathbf{F} \cdot \mathbf{v} = - F v = - A v^{\alpha+1},$

and the torque associated with the drag force reduces the angular momentum at a rate

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

For circular orbits, the energy can be expressed as a function of angular momentum:

$E(L) = - \frac{(GM)^2 m^3}{2 L^2}.$

Differentiating with respect to time gives

$\frac{dE}{dt} = \frac{dE}{dL} \frac{dL}{dt} = \frac{(GM)^2 m^3}{L^3} \frac{dL}{dt}.$

This ensures that the evolution of energy and angular momentum is compatible with the assumption of a nearly circular orbit.

Derivation

Expressing $E$ and $L$ as functions of $r$, one obtains

$E = -\frac{GM m}{2 r}, \quad L = m \sqrt{GM r}.$

Differentiating with respect to $r$ gives

$\frac{dE}{dr} = \frac{GM m}{2 r^2}, \quad \frac{dL}{dr} = \frac{m}{2} \sqrt{\frac{GM}{r}}.$

The rates of change due to drag are

$\frac{dE}{dt} = -A v^{\alpha+1} = - A (GM)^{\frac{\alpha+1}{2}} r^{-\frac{\alpha+1}{2}},$

$\frac{dL}{dt} = - A v^\alpha r = - A (GM)^{\frac{\alpha}{2}} r^{1 - \frac{\alpha}{2}}.$

Using the chain rule for circular orbits gives

$\frac{dE}{dt} = \frac{(GM)^2 m^3}{L^3} \frac{dL}{dt} = (GM)^{1/2} r^{-3/2} \frac{dL}{dt}.$

Substituting the expression for $dL/dt$ produces

$\frac{dE}{dt} = (GM)^{1/2} r^{-3/2} \left[- A (GM)^{\alpha/2} r^{1 - \alpha/2} \right] = - A (GM)^{(\alpha+1)/2} r^{- (\alpha + 1)/2},$

confirming consistency between the direct computation from the drag force and the chain rule relation.

The rate of change of the orbital radius follows from

$\frac{dE}{dt} = \frac{dE}{dr} \frac{dr}{dt},$

which gives

$\frac{GM m}{2 r^2} \frac{dr}{dt} = - A (GM)^{(\alpha+1)/2} r^{-(\alpha+1)/2}.$

Solving for $dr/dt$ yields

$\frac{dr}{dt} = - \frac{2 A (GM)^{(\alpha-1)/2}}{m} r^{(3-\alpha)/2}.$

This expression shows that the radial decay rate depends both on the constant $A$ and on the exponent $\alpha$.

To determine $\alpha$, one must impose the physical condition that the orbit remains nearly circular while slowly evolving under the drag force. The circularity condition requires that the ratio of the rate of change of angular momentum to the angular momentum itself equals the ratio of the rate of change of energy to the derivative of energy with respect to angular momentum. Equivalently, for the orbit to remain circular as it decays, the drag must satisfy the relation

$\frac{dE}{dt} = \frac{dE}{dL} \frac{dL}{dt}.$

Substituting the expressions for $dE/dt$ and $dL/dt$ gives

$- A v^{\alpha+1} = \frac{(GM)^2 m^3}{L^3} (- A v^\alpha r),$

which simplifies to

$v^{\alpha+1} = \frac{(GM)^2 m^3}{L^3} v^\alpha r.$

Using $L = m v r$ and cancelling common factors produces

$v^{\alpha+1} = \frac{(GM)^2 m^3}{(m v r)^3} v^\alpha r = \frac{(GM)^2 m^3}{m^3 v^3 r^3} v^\alpha r = \frac{(GM)^2}{v^3 r^3} v^\alpha r = \frac{(GM)^2}{v^{3-\alpha} r^{2}}.$

Equating powers of $v$ and $r$ separately, one finds

$v^{\alpha+1} = \frac{(GM)^2}{v^{3-\alpha} r^2} \quad \Rightarrow \quad v^{2 \alpha -2} = \frac{(GM)^2}{r^2}.$

Substituting $v^2 = GM/r$ gives

$(GM/r)^{\alpha - 1} = GM / r \quad \Rightarrow \quad \alpha - 1 = 1 \quad \Rightarrow \quad \alpha = 2.$

This derivation relies on the self-consistency of the orbital decay under the assumption of circular motion. The requirement that the energy and angular-momentum losses satisfy the circular-orbit relation uniquely determines $\alpha$.

Result

The exponent in the drag force that ensures a slowly decaying circular orbit is

$\alpha = 2.$

This value guarantees that the tangential drag reduces the satellite's energy and angular momentum in a manner compatible with circular orbits, allowing the orbital radius to shrink gradually without introducing eccentricity.

Sanity Checks

Dimensional analysis confirms consistency. For $\alpha = 2$, the units of $A$ are $[A] = [F]/[v^2] = (\mathrm{kg , m / s^2}) / (\mathrm{m/s})^2 = \mathrm{kg/m}$. The rate of radial decay scales as $dr/dt \sim r^{(3-\alpha)/2} = r^{1/2}$, showing slow decay for large $r$ when $A$ is small, consistent with the assumption of quasi-stationary circular motion. The energy and angular momentum evolution satisfy the exact circular-orbit relation $E(L)$, ensuring that the orbit remains circular while slowly contracting. Any other value of $\alpha$ would violate this condition, leading either to growth of eccentricity or inconsistent orbital evolution, confirming the physical necessity of $\alpha = 2$.