To determine the speed of the object as it hits the ground, we can use the conservation of energy. Since the object is falling freely under the influence of gravity, the sum of its gravitational potential energy and kinetic energy will remain constant throughout its descent.

Step 1: Initial and Final Energy States

1. Initial State (at height $h$):

   • The initial gravitational potential energy at a distance $(R + h)$ from the planet's center is: $U_i = -\frac{G M m}{R + h}$
   • The object is initially at rest, so its initial kinetic energy $K_i$ is zero: $K_i = 0$

   Therefore, the total initial energy $E_i$ is: $E_i = U_i + K_i = -\frac{G M m}{R + h}$

2. Final State (at the planet's surface):

   • When the object reaches the surface, it is at a distance $R$ from the center of the planet.
   • The final gravitational potential energy at the planet's surface is: $U_f = -\frac{G M m}{R}$
   • Let $v$ be the speed of the object when it hits the ground. Its final kinetic energy $K_f$ is: $K_f = \frac{1}{2} m v^2$

   Thus, the total final energy $E_f$ is: $E_f = U_f + K_f = -\frac{G M m}{R} + \frac{1}{2} m v^2$

Step 2: Apply Conservation of Energy

Since energy is conserved, we equate the initial and final total energies: $E_i = E_f$ $-\frac{G M m}{R + h} = -\frac{G M m}{R} + \frac{1}{2} m v^2$

Step 3: Solve for $v$

1. Rearrange to isolate $v^2$: $\frac{1}{2} m v^2 = \frac{G M m}{R} - \frac{G M m}{R + h}$

2. Factor out $G M m$ from the right-hand side: $\frac{1}{2} m v^2 = G M m \left( \frac{1}{R} - \frac{1}{R + h} \right)$

3. Cancel $m$ from both sides: $\frac{1}{2} v^2 = G M \left( \frac{1}{R} - \frac{1}{R + h} \right)$

4. Multiply by 2 to solve for $v^2$: $v^2 = 2 G M \left( \frac{1}{R} - \frac{1}{R + h} \right)$

5. Finally, take the square root to get $v$: $v = \sqrt{2 G M \left( \frac{1}{R} - \frac{1}{R + h} \right)}$

Final Answer

The speed $v$ of the object as it hits the ground is: $v = \sqrt{2 G M \left( \frac{1}{R} - \frac{1}{R + h} \right)}$

This expression gives the final speed of an object dropped from a height $h$ above the surface of a planet with mass $M$ and radius $R$, where $G$ is the gravitational constant.

Would you like more details on any part of this derivation?

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1. How would the expression change if air resistance were considered?
2. What would the final speed be if the object were dropped from infinity?
3. How does this formula simplify if $h \ll R$?
4. How would the impact speed change if the planet’s radius $R$ were larger?
5. What is the impact speed if the object is dropped from twice the planet's radius?

Tip: When solving problems involving gravitational potential, remember that gravitational potential energy depends on the inverse of the distance from the center of the mass creating the gravitational field.