Standing on the surface of a small spherical moon whose radius is \(6.30 \cdot 10^{4} \mathrm{~m}\) and whose mass is \(8.00 \cdot 10^{18} \mathrm{~kg}\) an astronaut throws a rock of mass 2.00 kg straight upward with an initial speed \(40.0 \mathrm{~m} / \mathrm{s}\). (This moon is too small to have an atmosphere.) What maximum height above the surface of the moon will the rock reach?

We have the Universal Law of Gravitation: \[F = G \frac{m_1 m_2}{r^2}\] And the potential energy is given by: \[U = -G \frac{m_1 m_2}{r}\] Here, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the moon and the rock, respectively, and \(r\) is the distance between their centers. The initial kinetic energy (at the moon's surface) can be calculated as: \[K_i = \frac{1}{2} m v^2\] where \(m\) is the mass of the rock, and \(v\) is its initial speed. Now, we will apply the conservation of mechanical energy, i.e., \[K_i + U_i = U_f\] where \(K_i\) is the initial kinetic energy, \(U_i\) is the initial gravitational potential energy, and \(U_f\) is the final gravitational potential energy at the maximum height.

When the astronaut throws the rock from the surface of the moon, the distance between the centers is equal to the moon's radius, \(R\). So, the initial gravitational potential energy is: \[U_i = -G \frac{M m}{R}\] Substitute the given values: \[U_i = - (\text{6.674}\times 10^{-11} \text{ Nm/kg}^2) \frac{(8.00 \times 10^{18} \text{ kg})(2.00 \text{ kg})}{(6.30 \times 10^4 \text{ m})}\] Calculate \(U_i\): \[U_i = -1.681 \times 10^7 \text{ J}\]

Using the initial speed given, we can calculate the initial kinetic energy: \[K_i = \frac{1}{2}m v^2\] Substituting the values: \[K_i = \frac{1}{2}(2.00 \text{ kg})(40.0 \text{ m/s})^2\] Calculate \(K_i\): \[K_i = 1.600 \times 10^3 \text{ J}\]

We know that \(K_i + U_i = U_f\), hence: \[1.600 \times 10^3 \text{ J} - 1.681 \times 10^7 \text{ J} = -G \frac{(8.00 \times 10^{18}\text{ kg})(2.00\text{ kg})}{R_f}\] Solving for \(R_f\): \[R_f = \frac{(8.00 \times 10^{18}\text{ kg})(2.00\text{ kg})(\text{6.674}\times 10^{-11} \text{ Nm/kg}^2)}{1.681 \times 10^7 \text{ J} - 1.600 \times 10^3 \text{ J}}\] Calculate \(R_f\): \[R_f = 6.326 \times 10^4 \text{ m}\]

Now that we have the final radius, we can calculate the maximum height the rock reaches above the moon's surface by subtracting the moon's radius from the final radius: \[H = R_f - R \] Substitute the values: \[H = (6.326 \times 10^4 \text{ m}) - (6.30 \times 10^4 \text{ m})\] Calculate \(H\): \[H = 260 \text{ m}\] The maximum height above the surface of the moon is 260 meters.