Question

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?

Short answer

Answer: The maximum height the rock will reach above the surface of the moon is approximately 12,460 meters.

Step-by-step solution

Identify the Initial and Final States

The initial state is when the rock is thrown with an initial speed of \(40.0 \mathrm{~m/s}\) from the surface of the moon. At this point, the rock has both kinetic and potential energy. The final state is when the rock reaches its maximum height, at which point it has zero velocity, and its kinetic energy is zero. At this moment, the rock has only potential energy.

Calculate the Initial Kinetic Energy

The initial kinetic energy of the rock can be calculated using the formula: \(KE = \frac{1}{2} mv^2\), where m is the mass of the rock and v is the initial velocity. \(KE = \frac{1}{2} (2.00 \mathrm{~kg}) (40.0 \mathrm{~m/s})^2 = 1600 \mathrm{~J}\)

Calculate the Initial Gravitational Potential Energy

The initial gravitational potential energy of the rock can be calculated using the formula: \(PE = -G\frac{Mm}{r}\), where G is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{N m^2/kg^2}\)), M is the mass of the moon, m is the mass of the rock, and r is the distance from the center of the moon to the rock (initially equal to the radius of the moon). \(PE = -6.674 \times 10^{-11} \mathrm{N m^2/kg^2} \frac{(8.00 \times 10^{18} \mathrm{~kg})(2.00 \mathrm{~kg})}{6.30 \times 10^4 \mathrm{~m}} = -1.687 \times 10^3 \mathrm{~J}\)

Apply the Conservation of Mechanical Energy

According to the conservation of mechanical energy, the initial total mechanical energy (kinetic + potential) is equal to the final potential energy. \(KE_{initial} + PE_{initial} = PE_{final}\) Replacing the values found in steps 2 and 3: \(1600 \mathrm{~J} - 1.687 \times 10^3 \mathrm{~J} = -G\frac{Mm}{r_{final}}\) Solve for \(r_{final}\): \(-6.674 \times 10^{-11} \mathrm{N m^2/kg^2} \frac{(8.00 \times 10^{18} \mathrm{~kg})(2.00 \mathrm{~kg})}{r_{final}} = -87 \mathrm{~J}\) \(r_{final} = 7.546 \times 10^4 \mathrm{~m}\)

Calculate the Maximum Height

To find the maximum height, subtract the initial radius of the moon from the final distance: \(height = r_{final} - r_{initial} = 7.546 \times 10^4 \mathrm{~m} - 6.30 \times 10^4 \mathrm{~m} = 1.246 \times 10^4 \mathrm{~m}\) The maximum height the rock will reach above the surface of the moon is approximately 12,460 meters.

Key concepts

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. On a small moon, like the one in our exercise, this energy is significant when calculating how high a thrown object will go. Given that the moon has a mass and radius, we use the formula \(PE = -G\frac{Mm}{r}\), where

• \(G\) is the gravitational constant \(6.674 \times 10^{-11} \mathrm{N m^2/kg^2}\),
• \(M\) is the mass of the moon,
• \(m\) is the mass of the rock,
• and \(r\) is the distance from the center of the moon to the rock.

Notice that the formula has a negative sign, reflecting the convention that gravitational potential energy is zero at an infinite distance and negative closer to the celestial body.
In the exercise, the initial gravitational potential energy is crucial in determining the highest point of the rock's trajectory. The rock’s potential energy at the beginning gives insight into how gravitational forces act upon it as it rises against gravity.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. For any object, it can be calculated using the formula \(KE = \frac{1}{2} mv^2\).In the problem, before the rock is thrown, it has kinetic energy due to its upward velocity of 40.0 m/s. This energy is stored in the fast-moving rock and will be entirely converted into potential energy at the peak of its flight.

• Note that the key here is the initial velocity - a higher speed means more kinetic energy and thus, a higher potential height.
• The kinetic energy will decrease as the rock moves higher until it reaches zero at the maximum point of the trajectory.

Understanding this transformation is crucial when studying energy conservation. As the rock ascends, its kinetic energy diminishes while gravitational potential energy increases. The dance between these two energies will dictate the rock’s motion in the void above.

Energy Conservation in Space

In the vastness of space, the conservation of mechanical energy principle becomes even more vital. Unlike on Earth, where air resistance and friction are present, in space environments like a small moon, these factors are absent. The total mechanical energy in such a scenario is preserved, meaning that the sum of kinetic and potential energy remains constant. Therefore,

• when the rock is thrown upwards, its kinetic energy begins to convert into gravitational potential energy,
• and upon reaching the maximum height, all the initial kinetic energy is stored as potential energy.

The equation \(KE_{initial} + PE_{initial} = PE_{final}\) highlights this principle. The rock’s journey is a perfect balance between the forces, illustrating the purity of physical laws in the space environment.
When you understand this energy transfer, it’s easier to comprehend how celestial motion operates, completely reliant on the conservation of energy principle. These principles are not just theoretical but practical, allowing predictions such as the maximum height achieved by the rock thrown from the moon's surface.