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The problem asks for the mass of a planet given specific conditions: a rock is released from rest at a height of 6.00 m above the surface, falls, and reaches a speed of 8.00 m/s just before striking the surface. The planet has a radius of \(6.00 \times 10^6 \, \text{m}\).

We can solve this problem using the principle of energy conservation. Here's the step-by-step solution:

### 1. **Energy Conservation Principle**:
We assume no air resistance, so mechanical energy is conserved. The initial potential energy of the rock when it's 6.00 m above the surface will be converted into kinetic energy just before impact.

- **Initial Energy**:
  - Potential Energy: \( U_i = \frac{GMm}{R} \), where \( M \) is the mass of the planet, \( m \) is the mass of the rock, \( G \) is the gravitational constant, and \( R = 6.00 \times 10^6 \, \text{m} \) is the radius of the planet.
  
- **Final Energy**:
  - Kinetic Energy: \( K_f = \frac{1}{2}mv^2 \), where \( v = 8.00 \, \text{m/s} \).

### 2. **Work-Energy Relation**:
For the rock dropped from a small height, the change in gravitational potential energy (from height 6.00 m to the surface) equals the final kinetic energy. So:

\[
\Delta U = K_f
\]

The change in potential energy for a small height \( h \) near the surface can be approximated as:

\[
\Delta U \approx mgh
\]

Where \( g \) is the acceleration due to gravity on the planet's surface and \( h = 6.00 \, \text{m} \). The kinetic energy is:

\[
\frac{1}{2}mv^2
\]

Equating the two:

\[
mgh = \frac{1}{2}mv^2
\]

The mass \( m \) cancels out:

\[
gh = \frac{1}{2}v^2
\]

### 3. **Solve for \( g \)**:
Now solve for \( g \), the gravitational acceleration:

\[
g = \frac{v^2}{2h} = \frac{(8.00)^2}{2 \times 6.00} = \frac{64.0}{12.0} = 5.33 \, \text{m/s}^2
\]

### 4. **Gravitational Acceleration Formula**:
The gravitational acceleration is related to the mass of the planet by Newton's law of gravitation:

\[
g = \frac{GM}{R^2}
\]

Substitute the known values (\( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( R = 6.00 \times 10^6 \, \text{m} \), and \( g = 5.33 \, \text{m/s}^2 \)):

\[
5.33 = \frac{(6.67 \times 10^{-11})M}{(6.00 \times 10^6)^2}
\]

Simplify:

\[
5.33 = \frac{6.67 \times 10^{-11} M}{3.6 \times 10^{13}}
\]

Now, solve for \( M \):

\[
M = \frac{5.33 \times 3.6 \times 10^{13}}{6.67 \times 10^{-11}} = 2.88 \times 10^{24} \, \text{kg}
\]

This is approximately \( 2.9 \times 10^{24} \, \text{kg} \), which matches option **(b)**.

### Conclusion:
The mass of the planet is \( \boxed{2.9 \times 10^{24} \, \text{kg}} \).

---

Do you have any questions or would you like further details on any part of this solution?

### Related Questions:
1. How would the result change if the radius of the planet were doubled?
2. How does air resistance affect the conservation of mechanical energy in real-world situations?
3. What if the rock had been dropped from a height significantly greater than 6 m? How would we approach the problem?
4. How can you derive the formula for gravitational acceleration from Newton’s law of gravitation?
5. Why does the mass of the rock cancel out in this problem?

### Tip:
When solving problems with gravitational forces, keep in mind that energy conservation often simplifies calculations, especially near the surface where the gravitational field can be approximated as uniform.

You are exploring a planet. You release a small rock from rest at a height of 6.00 m above the surface of the planet and find that its speed is 8.00 m/s just before it strikes the surface. If the planet has a radius of 6.00×10^6 m, what is its mass? Assume that the planet has spherical symmetry.

Explore how to determine the mass of a planet by applying energy conservation principles and Newton's law of gravitation. The problem involves calculating gravitational potential energy and final kinetic energy when a rock is released from a small height above the surface of a planet. This is a suitable topic for college-level physics students.

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