Step 1: Recall the formula for total mechanical energy of a satellite.
For a satellite of mass $m$ in a circular orbit of radius $r$ around Earth (mass $M$), the total mechanical energy is the sum of kinetic and potential energy: \[ E = -\frac{GMm}{2r} \] The negative sign indicates that the satellite is gravitationally bound. The kinetic energy is $+GMm/(2r)$ and potential energy is $-GMm/r$, so the total is $-GMm/(2r)$.
Step 2: Calculate the initial total energy.
Let the initial orbital radius be $r$. The initial total energy is: \[ E_i = -\frac{GMm}{2r} \]
Step 3: Calculate the final total energy after the radius is halved.
The new orbital radius is $r_f = r/2$. Substituting: \[ E_f = -\frac{GMm}{2(r/2)} = -\frac{GMm}{r} \] Comparing magnitudes: $|E_f| = GMm/r$ while $|E_i| = GMm/(2r)$, so $|E_f| = 2|E_i|$. The total energy magnitude doubles when the orbital radius is halved.
Step 4: Calculate the change in the magnitude of total energy.
\[ \Delta|E| = |E_f| - |E_i| = \frac{GMm}{r} - \frac{GMm}{2r} = \frac{GMm}{2r} = |E_i| \] The change in magnitude equals the initial magnitude.
Step 5: Express the percentage change.
\[ \% \text{ change} = \frac{|E_f| - |E_i|}{|E_i|} \times 100 = \frac{|E_i|}{|E_i|} \times 100 = 100\% \]
Step 6: Interpret the physical meaning.
When the satellite's orbital radius is reduced to half, its total energy magnitude doubles. The satellite is more tightly bound to Earth (more negative total energy). In practice, to reduce the orbital radius, the satellite loses energy (e.g., by firing retro-rockets), and the energy released equals the initial total energy magnitude, corresponding to a 100% change. \[ \boxed{100\%} \]