Problem 245 - Calculus Volume 1

Question

For the following exercises, find the work done. [T] The force of gravity on a mass \(m\) is \(F=-\left((G M m) / x^{2}\right)\) newtons. For a rocket of mass \(m=1000 \mathrm{kg}, \quad\) compute the work to lift the rocket from \(x=6400\) to \(x=6500\) km.(Note:\(G=6 \times 10^{-17} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\) and \(M=6 \times 10^{24} \mathrm{kg} .\)

Step-by-Step Solution

Answer

The work done is approximately 9.6 \times 10^8 joules.

Step 1: Understand the Formula for Work Done

Work done to move an object against a variable force can be found using the integral of the force over the distance it moves. In this problem, integrate the gravitational force function, which is given as \[ F(x) = -\frac{G M m}{x^2} \] from \( x_1 = 6400 \) km to \( x_2 = 6500 \) km.

Step 2: Convert Units

Convert distances from kilometers to meters to use SI units consistently. This means \[ x_1 = 6400 \text{ km} = 6.4 \times 10^6 \text{ m} \] and\[ x_2 = 6500 \text{ km} = 6.5 \times 10^6 \text{ m}. \]

Step 3: Set Up the Integral

The work done \( W \) can be calculated as \[ W = -\int_{x_1}^{x_2} \frac{G M m}{x^2} \, dx. \] Substitute the values of \( G \), \( M \), and \( m \). This gives\[ W = -\int_{6.4 \times 10^6}^{6.5 \times 10^6} \frac{6 \times 10^{-17} \times 6 \times 10^{24} \times 1000}{x^2} \, dx. \]

Step 4: Evaluate the Integral

Perform the integration. The integral of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \), leading to\[ W = -G M m \left[ -\frac{1}{x} \right]_{6.4 \times 10^6}^{6.5 \times 10^6}. \]Simplify to get\[ W = G M m \left( \frac{1}{6.4 \times 10^6} - \frac{1}{6.5 \times 10^6} \right). \]

Step 5: Calculate the Work Done

Substitute the constants and compute:\[ W = (6 \times 10^{-17})(6 \times 10^{24})(1000) \left( \frac{1}{6.4 \times 10^6} - \frac{1}{6.5 \times 10^6} \right). \]Calculate the numerical difference in fractions, and multiply by the other constants to find the work in joules.

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept in physics, describing the attractive force that acts between two masses. It is given by Newton's law of universal gravitation and is mathematically expressed as \[ F = -\frac{G M m}{x^2} \]where:

• \( G \) is the gravitational constant.
• \( M \) and \( m \) are the masses of the two objects.
• \( x \) is the distance between the centers of the two masses.

The negative sign indicates that the force is attractive, pulling the two masses towards each other. Gravitational force decreases with the square of the distance, meaning it becomes weaker as the objects are farther apart.
This understanding is critical when calculating work done against gravitational force, especially in space travel, where distances are significant.

Integration in Calculus

Integration in calculus is a powerful tool to compute quantities that accumulate over a given interval, like work done by a force. When dealing with variable forces, like gravitational force which changes with distance, integration becomes essential.
Integration allows us to sum up infinitely small quantities to find a total. In our rocket example, the gravitational force is a function of distance, and thus, to calculate the work done against this force, we need to integrate it over a specified distance. The formula for work done involves the integral of the force function:\[ W = \int_{x_1}^{x_2} F(x) \, dx \]This gives us the total work required to move an object from point \( x_1 \) to \( x_2 \) against the gravitational pull.

Unit Conversion

Unit conversion plays a crucial role in ensuring accurate calculations in physics. Consistent units help avoid errors and represent quantities in an internationally recognized system. In our problem, we converted distances from kilometers (km) to meters (m) since the International System of Units (SI) uses meters as the standard unit for distance.
To convert kilometers to meters, simply multiply by 1000. For instance:

• \( 6400 \text{ km} = 6.4 \times 10^6 \text{ m} \)
• \( 6500 \text{ km} = 6.5 \times 10^6 \text{ m} \)

Accuracy in these conversions is vital as any errors can greatly affect the outcome, especially in scientific calculations involving large or small quantities.

Definite Integral

The concept of a definite integral is central when calculating the work done over a specific interval. A definite integral involves integration with defined upper and lower limits, providing a precise total value over that interval.
When dealing with physical problems like work done by or against a force, the definite integral is utilized to calculate the total accumulation of that force over a distance.
In our exercise, the definite integral:\[ W = -\int_{6.4 \times 10^6}^{6.5 \times 10^6} \frac{6 \times 10^{-17} \times 6 \times 10^{24} \times 1000}{x^2} \, dx \]calculates the total work done to move the rocket between the specific distances of 6400 km and 6500 km. This provides the exact amount of energy required to overcome the gravitational pull within these boundaries.