Solution 5
Derivation:
Total mechanical energy of a satellite of mass $m$ in a circular orbit of radius $r$ is:
$$ E = -\frac{GMm}{2r} $$The rate of loss of energy is equal to the power dissipated by the retarding force $F$. Since the force acts opposite to the velocity $v$:
$$ \frac{dE}{dt} = \vec{F} \cdot \vec{v} = -Fv $$Differentiating the energy equation with respect to time:
$$ \frac{dE}{dt} = \frac{d}{dt} \left( -\frac{GMm}{2r} \right) = \frac{GMm}{2r^2} \frac{dr}{dt} $$Equating the two expressions for power:
$$ \frac{GMm}{2r^2} \frac{dr}{dt} = -Fv $$ $$ \frac{dr}{dt} = -\frac{2Fv r^2}{GMm} $$For a circular orbit, orbital velocity is given by $v = \sqrt{\frac{GM}{r}}$, which implies $r = \frac{GM}{v^2}$. Substituting $r$ into the equation:
$$ \frac{dr}{dt} = -\frac{2Fv}{GMm} \left( \frac{GM}{v^2} \right)^2 = -\frac{2Fv}{GMm} \frac{G^2 M^2}{v^4} = -\frac{2F GM}{m v^3} $$Since the problem states “very low altitude”, we assume $r \approx R$ (Radius of Earth). The acceleration due to gravity near the surface is $g = \frac{GM}{R^2}$. Also $v \approx \sqrt{gR}$.
Alternatively, using $GM = gR^2$ and $v^2 = gR$:
$$ \frac{dr}{dt} = -\frac{2F (v^2 r)}{m v^2} \frac{1}{r} \dots \text{(simplifying directly)} $$Let’s use the simplest substitution form $GM = v^2 r$ in the first derivative step:
$$ \frac{dr}{dt} = -\frac{2Fv r^2}{(v^2 r)m} = -\frac{2Fr}{mv} $$For a low orbit, $r \approx R$. We also know $mg = \frac{mv^2}{R}$ (Centripetal force provided by gravity). Thus $R = \frac{v^2}{g}$. Substituting this $R$ for $r$:
$$ \frac{dr}{dt} = -\frac{2F}{mv} \left( \frac{v^2}{g} \right) = -\frac{2Fv}{mg} $$Answer: $\frac{dr}{dt} = -\frac{2Fv}{mg}$