Problem 245 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The work done to lift the rocket is approximately 609.23 Joules.

Step 1: Understand the Problem

The problem requires us to calculate the work done to lift a rocket from 6400 km to 6500 km against the gravitational force. The force of gravity on a mass is given by the equation \( F = -\left( \frac{G M m}{x^2} \right) \).

Step 2: Set up the Work Integral

Work done by a force is the integral of the force over a distance. Therefore, the work \( W \) needed to lift the rocket from \( x=6400 \) km to \( x=6500 \) km is given by the integral:\[W = \int_{6400 \times 10^3}^{6500 \times 10^3} \left( \frac{G M m}{x^2} \right) \, dx\]We omit the negative sign as we are calculating the work done against gravity.

Step 3: Substitute Known Values

Substitute the values into the integral:\[W = \int_{6400 \times 10^3}^{6500 \times 10^3} \left( \frac{6 \times 10^{-17} \cdot 6 \times 10^{24} \cdot 1000}{x^2} \right) \, dx\]Replace the constants \( G \), \( M \), and \( m \) with their respective values.

Step 4: Simplify and Compute the Integral

Simplify the integral:\[W = 36 \times 10^8 \cdot \int_{6400 \times 10^3}^{6500 \times 10^3} \frac{1}{x^2} \, dx\]The integral \( \int \frac{1}{x^2} \, dx \) evaluates to \( -\frac{1}{x} \).

Step 5: Evaluate the Definite Integral

Substitute the limits into the antiderivative:\[W = 36 \times 10^8 \left( -\frac{1}{6500 \times 10^3} + \frac{1}{6400 \times 10^3} \right)\]Calculate the result to find the work done.

Step 6: Calculate Numerical Value

Perform the arithmetic calculations:\[W = 36 \times 10^8 \left( \frac{1}{6400 \times 10^3} - \frac{1}{6500 \times 10^3} \right)\]\[W = 36 \times 10^8 \left(\frac{5.7692308 \times 10^{-8} - 5.6 \times 10^{-8}}{1} \right)\]\[W = 36 \times 10^8 \times 0.1692308 \times 10^{-8}\]\[W \approx 609.23 \text{ J}\]Therefore, the work done is approximately 609.23 Joules.

Key Concepts

Gravitational force

Gravitational force is a fundamental natural phenomenon that attracts every object with mass to every other object with mass in the universe. This force is why when you drop something, it falls to the ground. The equation for gravitational force between two masses is given by Newton's Law of Universal Gravitation:

• \( F = -\left( \frac{G M m}{x^2} \right) \)

Here:

• \( G \) is the gravitational constant (\( 6 \times 10^{-17} \, \text{N m}^2/ \text{kg}^2 \))
• \( M \) is the mass of the larger object (e.g., Earth)
• \( m \) is the mass of the smaller object (e.g., a rocket)
• \( x \) is the distance between the centers of the two masses

Gravity's strength decreases with the square of the distance, showing why the force is stronger closer to the Earth. Understanding gravitational force is crucial in fields like physics and engineering since it affects everything from orbiting satellites to falling apples.
In our specific exercise, we're finding how much work it takes to lift a rocket against this gravitational pull.

Integral calculus

Integral calculus is a branch of mathematical analysis dealing with integrals and their properties. It is used to find quantities like areas, volumes, and in the context of physics, work done by a force.
When you hear "integral calculus," two important operations come to mind:

• Indefinite Integrals: Used to find antiderivatives - they reverse differentiation. They give us a function or functions that represent all antiderivatives.
• Definite Integrals: Used to find a specific numerical value for the area under a curve or the total accumulation of a rate of change over an interval.

For example, when calculating work, integral calculus helps us sum up tiny amounts of work done over small distances into a total amount of work done over a larger distance. This is exactly how we calculated the rocket's work against Earth's gravity, finding the integral of the gravitational force over the path.

Definite integral

The definite integral is a tool in calculus that allows us to calculate the exact value of the integral over a specified interval. It is denoted as:

• \[ \int_{a}^{b} f(x) \, dx \]

In essence, it adds up (or integrates) an infinite number of infinitesimally small quantities over an interval, giving us a total.

• \( a \) and \( b \) are the lower and upper limits of the interval.

In our exercise, we used a definite integral to find the work done in lifting the rocket from 6400 km to 6500 km. This means integrating the gravitational force equation across these bounds:

• \[ \int_{6400 \times 10^3}^{6500 \times 10^3} \left( \frac{G M m}{x^2} \right) \, dx \]

By evaluating the definite integral, you solve the integral expression and then substitute the upper and lower limits to get the work done.
It's a precise way due to how the definite integral accounts for all values in the interval, summing infinitely small contributions to the total work.