A 1200 kg satellite orbits 420 km up. Over some weeks, the thin air takes 2.0 × 10⁸ J of its energy.

Step 1. Before. $E_1 = -\dfrac{4.014 \times 10^{14}\ \text{m}^3/\text{s}^2 \times 1200\ \text{kg}}{2 \times 6.82 \times 10^{6}\ \text{m}} = -3.5314 \times 10^{10}\ \text{J}$.

Step 2. After. $E_2 = -3.5514 \times 10^{10}\ \text{J}$, so $r_2 = \dfrac{GMm}{2|E_2|} = 6.7816 \times 10^{6}\ \text{m}$: a height of $381.6\ \text{km}$.

Step 3. Kinetic energy. $K = -E$ rises by $2.0 \times 10^{8}\ \text{J}$.