Irodov 1.227: Drag Spirals a Moon Satellite Down – Amit & Deepali Sinha Foundation

Problem Irodov 1.227

A satellite of mass $m$ circles the Moon (mass $M$ , radius $R$ ) in a circular orbit of radius $r=\eta R$ . A weak drag force of magnitude $F=\alpha v^2$ , with $\alpha$ a known constant and $v$ the satellite’s speed, continually removes energy from the orbit. Find the time $\tau$ until the satellite spirals down onto the Moon’s surface.

The satellite m on its present circular orbit of radius η R about the Moon M, at speed v. The decay ends at the Moon's surface, so the final orbit R lies on the rim itself. — The satellite $m$ on its present circular orbit of radius $\eta R$ about the Moon $M$ , at speed $\vect v$ . The decay ends at the Moon’s surface, so the final orbit $R$ lies on the rim itself.

Given

orbit radius $r=\eta R$
resistance $F=\alpha v^2$
satellite mass $m$

Find

the time

\tau

until the satellite falls onto the Moon’s surface

Solution

Step 1Energy of a circular orbit at radius

r

At any instant the orbit is very nearly circular at its current radius $r$ , so gravity supplies the centripetal force there; combining this with the kinetic and potential energy at that radius gives the total mechanical energy as a function of $r$ alone.

\begin{align*} \frac{\gamma Mm}{r^2} = \frac{mv^2}{r} &\;\Longrightarrow\; v^2 = \frac{\gamma M}{r}, \tag{1} \\ E(r) = \frac12 mv^2 - \frac{\gamma Mm}{r} &= \frac{\gamma Mm}{2r}-\frac{\gamma Mm}{r}. \tag{2}\end{align*}

E(r) = -\frac{\gamma Mm}{2r}

(3)

Step 2Rate of energy loss to drag

The drag force opposes the motion, so it removes mechanical energy at the rate it does work against the satellite, $Fv$ ; writing $F=\alpha v^2$ turns this into a pure power of $v$ .

-\frac{dE}{dt} = Fv = \alpha v^3. \tag{4}

Step 3Convert the energy loss into a rate of change of

r

Differentiating (3) with respect to $r$ , then using the chain rule $dE/dt = (dE/dr)(dr/dt)$ together with (4), isolates $dr/dt$ ; substituting $v^2=\gamma M/r$ from (1) leaves an equation in $r$ and $t$ alone.

\begin{align*} \frac{dE}{dr} = \frac{\gamma Mm}{2r^2} &\;\Longrightarrow\; \frac{\gamma Mm}{2r^2}\frac{dr}{dt} = -\alpha v^3, \\ \frac{dr}{dt} &= -\frac{2\alpha v^3 r^2}{\gamma Mm} = -\frac{2\alpha r^2}{\gamma Mm}\left(\frac{\gamma M}{r}\right)^{3/2}. \tag{5}\end{align*}

\frac{dr}{dt} = -\frac{2\alpha\sqrt{\gamma M}}{m}\,r^{1/2}

(6)

Step 4Integrate from the present orbit down to the surface

Separating $r$ and $t$ in (6) and integrating from $r=\eta R$ at $t=0$ to $r=R$ at $t=\tau$ , when the satellite reaches the surface, gives $\tau$ .

\begin{align*} \int_{\eta R}^{R} r^{-1/2}\,dr &= -\frac{2\alpha\sqrt{\gamma M}}{m}\int_0^\tau dt \\ 2\sqrt{R}-2\sqrt{\eta R} &= -\frac{2\alpha\sqrt{\gamma M}}{m}\,\tau \\ \tau &= \frac{m\sqrt{R}\,(\sqrt\eta-1)}{\alpha\sqrt{\gamma M}} = \frac{m(\sqrt\eta-1)}{\alpha\sqrt{\gamma M/R}}. \tag{7}\end{align*}

Since $\gamma M/R^2$ is the Moon’s own surface gravity $g$ , the ratio $\gamma M/R$ equals $gR$ .

\tau \approx \frac{m(\sqrt\eta-1)}{\alpha\sqrt{gR}}

(8)

Check

Dimensionally $\alpha$ carries units kg/m, since $F=\alpha v^2$ ; multiplying by $\sqrt{gR}$ , a speed, gives kg/s, and dividing the mass $m$ by that leaves a time, as required. As $\eta\to1$ (the satellite already grazing the surface) the boxed $\tau\to0$ , consistent with no distance left to spiral through.

Remark

The trick worth keeping is converting a $t$ -integral into an $r$ -integral by writing $E$ and $v$ as functions of $r$ alone (Step 1), valid whenever the perturbation is weak enough that the orbit stays nearly circular at every instant; the same reduction works for any drag law $F(v)$ , not only $\alpha v^2$ , since it only ever used $E(r)$ and $-dE/dt=Fv$ .