1. The problem asks us to find the gravitational potential energy (PE) of a tungsten rod in low-Earth orbit at an altitude of 2000 km.\n2. The formula for gravitational potential energy relative to Earth is $$PE = -\frac{GMm}{r}$$ where:\n- $G$ is the gravitational constant $6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2}$\n- $M$ is the mass of Earth $5.972 \times 10^{24} \, kg$\n- $m$ is the mass of the rod $8,322 \, kg$\n- $r$ is the distance from Earth's center in meters\n3. Earth's radius $R_E$ is approximately $6,371 \, km = 6.371 \times 10^{6} \, m$. The altitude is $2,000 \, km = 2.0 \times 10^{6} \, m$.\n4. So, $r = R_E + \text{altitude} = 6.371 \times 10^{6} + 2.0 \times 10^{6} = 8.371 \times 10^{6} \, m$.\n5. Substitute values:$$PE = -\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 8,322}{8.371 \times 10^{6}}$$\n6. Calculate numerator: $$6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 8,322 = 3.316 \times 10^{18}$$ approximately.\n7. Divide by $r$: $$\frac{3.316 \times 10^{18}}{8.371 \times 10^{6}} = 3.960 \times 10^{11}$$ approximately.\n8. Thus, $$PE = -3.960 \times 10^{11} \, J$$ (joules).\n9. Convert joules to gigajoules (GJ): $$1 \, GJ = 10^{9} \, J$$ so $$PE = -396.00 \, GJ$$\n10. The negative sign indicates the energy is gravitational potential energy bound to Earth. The magnitude is about 396.00 GJ.\n\nFinal answer: The tungsten rod has approximately $396.00$ GJ of gravitational potential energy in low-Earth orbit at 2000 km altitude.