Solution

1. Qualitative explanation:
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.

2. Quantitative explanation:
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: $E = -E_k$.

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:
$$m \frac{dv}{dt} = F_{\text{drag}} \implies a_{\tau} = \frac{F_{\text{drag}}}{m}$$