on On of your summer lunar space camp activities is to launch a $1130 \mathrm{~kg}$ rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of $207 \mathrm{~km}$. What gain $\Delta U$ in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are $7.36 \times 10^{22} \mathrm{~kg}$ and $1740 \mathrm{~km}$, respectively. \[ \Delta U= \]

Step 1 ：The gravitational potential energy is given by the formula: \(U = - \frac{GMm}{r}\) where G is the gravitational constant, M is the mass of the celestial body (in this case, the Moon), m is the mass of the object (in this case, the rocket), and r is the distance from the center of the celestial body.

Step 2 ：The change in gravitational potential energy, ΔU, is given by the difference in potential energy at the final and initial positions. In this case, the initial position is on the surface of the Moon (r = radius of the Moon) and the final position is at an altitude of 207 km above the surface of the Moon (r = radius of the Moon + 207 km).

Step 3 ：So, we need to calculate: \(\Delta U = U_{final} - U_{initial} = - \frac{GMm}{r_{final}} - (- \frac{GMm}{r_{initial}})\)

Step 4 ：Let's plug in the given values: M = 7.36e+22 kg, m = 1130 kg, r_initial = 1740000 m, r_final = 1947000 m.

Step 5 ：Calculate the initial potential energy: \(U_{initial} = - \frac{GMm}{r_{initial}} = -3190161967.816092 J\)

Step 6 ：Calculate the final potential energy: \(U_{final} = - \frac{GMm}{r_{final}} = -2850992205.4442735 J\)

Step 7 ：Calculate the change in potential energy: \(\Delta U = U_{final} - U_{initial} = 339169762.37181854 J\)

Step 8 ：The gain in gravitational potential energy accomplished by the launch is approximately 339169762.37 J. Therefore, \(\Delta U = \boxed{339169762.37 J}\)