Problem 33 - Single Variable Calculus

Question

A lunar landing module with a weight of \(20,000 \mathrm{lb}\), as measured on the earth, is to be launched vertically upward from the surface of the moon to a height of 20 mi. Taking the radius of the moon to be \(1100 \mathrm{mi}\) and its gravitational force to be one sixth that of the earth, find the work required to accomplish the task. Hint: See Exercise 32 .

Step-by-Step Solution

Answer

The required work to launch the lunar landing module is given by: \[W = -G M_{moon} \frac{W_{earth}}{g_{earth}} \left(\frac{1}{r_{final}} - \frac{1}{r_{initial}}\right)\] Substitute the given values and constants, and convert units if necessary, to find the answer.

Step 1: Calculate the mass of the lunar landing module

First, we need to find the mass of the lunar landing module. We know that on the earth, the weight of the module is 20,000 lb. The weight of an object is the product of its mass and the gravitational acceleration: \[W_{earth} = m \cdot g_{earth}\] where \(W_{earth}\) is the weight on Earth (20,000 lb), \(m\) is the mass of the module, and \(g_{earth}\) is the gravitational acceleration on Earth. Since the gravitational force on the moon is one sixth that of the earth, we have: \[g_{moon} = \frac{1}{6} g_{earth}\] Now we can calculate the weight of the module on the moon: \[W_{moon} = m \cdot g_{moon} = m \cdot \left(\frac{1}{6} g_{earth}\right)\] We can then solve for the mass \(m\) by using the given weight on Earth: \[m = \frac{W_{earth}}{g_{earth}}\] Note that the mass remains constant whether the module is on Earth or the moon.

Step 2: Determine the change in gravitational potential energy

Now we need to find the change in gravitational potential energy from the initial state (on the moon's surface) to the final state (20 mi above the moon's surface). The gravitational potential energy is given by: \[U = -G \frac{M_{moon} \cdot m}{r}\] where \(U\) is the gravitational potential energy, \(G\) is the universal gravitational constant, \(M_{moon}\) is the mass of the moon, \(m\) is the mass of the module, and \(r\) is the distance from the center of the moon. Initially, the module is on the moon's surface, so the distance from the center is the moon's radius: \[r_{initial} = 1100 \, \mathrm{mi}\] Finally, the module is 20 mi above the moon's surface: \[r_{final} = r_{initial} + 20 \, \mathrm{mi} = 1120 \, \mathrm{mi}\] The change in gravitational potential energy is the difference between the final and initial energies: \[\Delta U = U_{final} - U_{initial} = -G M_{moon} m \left(\frac{1}{r_{final}} - \frac{1}{r_{initial}}\right)\]

Step 3: Calculate the required work

The work required to launch the module is equal to the change in gravitational potential energy: \[W = \Delta U\] Now we have all the information to find the work required to accomplish the task: \[W = -G M_{moon} \frac{W_{earth}}{g_{earth}} \left(\frac{1}{r_{final}} - \frac{1}{r_{initial}}\right)\] We can now plug in the given values and constants to find the answer. Please note that you may need to convert the units to maintain consistency.

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is an important concept to grasp when dealing with objects being lifted above a celestial body's surface. This energy is the potential energy an object possesses due to its position in a gravitational field. For the lunar module problem, we need to calculate the change in gravitational potential energy as the module moves from the surface of the moon to a point 20 miles above it.

The formula used in the solution for gravitational potential energy involves the gravitational constant (G), the mass of the moon, the mass of the module, and the module's distance from the center of the moon:

• \[U = -G \frac{M_{moon} \cdot m}{r}\]

The negative sign accounts for gravitational attraction. The work done is equivalent to the change in this potential energy as the module moves to the higher position. Therefore, understanding potential energy is crucial for solving problems involving elevation in a gravitational field.

Gravitational Force

Gravitational force is a fundamental force that attracts any two masses. It's what keeps us rooted to Earth and allows celestial bodies to interact with each other. For this exercise, the gravitational force of the moon plays a vital role. To solve the problem, it's essential to understand that the moon's gravitational force is much weaker than Earth's—only about one sixth.

This is why objects weigh less on the moon. Even though they have the same mass, the force pulling them down is less intense:

• \[g_{moon} = \frac{1}{6} g_{earth}\]

This formula indicates how lunar gravity affects weight without affecting mass. Gravitational force is an interplay of mass and distance, forming the foundation for understanding more complex gravitational interactions.

Lunar Gravity

Lunar gravity can be fascinating because it provides an excellent example of how gravity varies across different celestial bodies. Since the moon has less mass than Earth, its gravitational pull is significantly weaker. This causes objects on the moon to weigh less but retain the same mass.

In our exercise, understanding lunar gravity helps in determining how much work is necessary to move the lunar module. As gravity provides the acceleration on the moon, it appears in weight calculations as:

• \[W_{moon} = m \cdot g_{moon}\]

The experience of lunar gravity explains why astronauts can hop and cover distances more easily on the moon compared to Earth. It's a critical detail for missions that consider fuel, energy, and mechanical work for movement.

Weight and Mass Conversion

Weight and mass are often confused but are distinct concepts. Mass is the amount of matter in an object and remains constant regardless of location. Weight, however, is the gravitational force acting on that mass, and it varies depending on the gravitational field you're in.

In our example, the lunar module weighs 20,000 lb on Earth. When converting the weight for lunar conditions, the process is simple: we adjust for the moon's lower gravity. The weight on the moon can be calculated using:

• \[W_{moon} = m \cdot g_{moon}\]

Mass is calculated using the known weight on Earth and the gravitational constant. This conversion is essential for carrying out correct calculations for movement in different gravitational fields. Remember, even though an object's weight changes, its mass stays the same because it is independent of gravity.