Problem 32 - Single Variable Calculus

Question

Show that the work \(W\) required to launch a rocket of mass \(m\) from the ground vertically upward to a height \(h\) is given by the formula $$ W=\frac{m g R h}{R+h} $$ Hint: Use Newton's Law of Gravitation given in Exercise 31 and follow these steps: (i) Let \(m\) and \(M\) denote the mass of the rocket and the earth, respectively, so that \(F=G m M / r^{2}\), where \(R \leq r \leq R+h\). At \(r=R\) the force will be the weight of the rocket, that is, \(m g=G m M / R^{2} .\) Therefore, \(G=g R^{2} / M\), so \(F=m g R^{2} / r^{2}\). (ii) \(W=\int_{R}^{R+h} F d r\). where \(R\) is the radius of the earth.

Step-by-Step Solution

Answer

To find the work required to launch a rocket of mass \(m\) vertically upward to a height \(h\), we first derive an expression for the gravitational force acting on the rocket as a function of its distance from the center of the Earth, \(F(r)=\frac{mgR^2}{r^2}\). Then, we integrate this force expression between the Earth's surface (\(r=R\)) and the desired height (\(r=R+h\)), giving us the work \(W=\int_{R}^{R+h} F(r) dr\). After evaluating the integral and simplifying the expression, we find that the work required to launch the rocket is given by the formula: \(W=\frac{mgRh}{R+h}\).

Step 1: (i) Force expression for the rocket

First, we need to find the expression for the gravitational force acting on the rocket. The rocket has mass \(m\) and the earth has mass \(M\). The gravitational force acting on the rocket at a distance \(r\) from the center of the earth is given by: $$ F = \frac{GmM}{r^2} $$ At the surface of the Earth, \(r=R\), the force will be equal to the weight of the rocket, \(mg\): $$ m g = \frac{GmM}{R^2} \implies G = \frac{gR^2}{M} $$ Now, we can rewrite the force acting on the rocket as a function of \(r\): $$ F(r) = \frac{mgR^2}{r^2} $$

Step 2: (ii) Calculate the work \(W\)

To find the total work required to take the rocket from the ground to a height \(h\), we need to integrate the force expression over the distance traveled: $$ W = \int_{R}^{R+h} F(r) dr $$ Now, substitute the force expression into the integral: $$ W = \int_{R}^{R+h} \frac{mgR^2}{r^2} dr $$ To integrate this expression, we can apply the integral of the form: $$ \int \frac{1}{r^2} dr = -\frac{1}{r} $$ So, by following the integration, we get: $$ W = mgR^2 \Biggl(-\frac{1}{r} \Biggr|_{R}^{R+h} $$ Now, evaluate the integral expression between the limits: $$ W = mgR^2 \left( -\frac{1}{R+h} + \frac{1}{R} \right) $$ Finally, simplify the expression to find the work required to launch the rocket to a height \(h\): $$ W = \frac{mgRh}{R+h} $$

Key Concepts

Gravitational Force

Gravitational force is a fundamental interaction in nature, attracting two bodies towards each other. This force is what keeps planets in orbit around the sun and what causes an apple to fall to the ground. It's directly proportional to the product of the two masses involved and inversely proportional to the square of the distance between their centers.

Gravitational force can be expressed using the formula:

• \( F = \frac{GmM}{r^2} \)

Here,

• \( F \) is the gravitational force.
• \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \) m\(^3\) kg\(^{-1}\) s\(^{-2}\)).
• \( m \) and \( M \) are the masses of the objects.
• \( r \) is the distance between the centers of the two masses.

The strength of gravity diminishes with distance, which is essential when calculating forces on bodies like rockets moving away from planets. Understanding this force is crucial in physics problem solving, especially in scenarios involving celestial bodies. It allows us to predict the motion of objects under the influence of gravity.

Newton's Law of Gravitation

Sir Isaac Newton's formulation of the law of gravitation was groundbreaking and forms the foundation of classical physics. According to Newton's Law of Gravitation, every point mass attracts every other point mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between the two points.

Newton’s Law can be expressed as:

• \( F = \frac{GmM}{r^2} \)

This equation helps us calculate the gravitational force and, implicitly, tells us that:

• The force decreases with increasing distance between masses (\( r \)).
• A larger mass (\( m \) or \( M \)) results in a greater force.

In practical applications, like launching rockets, Newton's law aids in calculating how much work (energy transferred by force) is necessary to move an object against gravitational attraction. Adjustments such as the change in force with distance due to the variable factor \( r \) relate directly to integration in calculus, another core concept in physics calculations and solutions.

Integration in Calculus

Integration is an essential concept in calculus that helps in finding quantities like area under a curve, total accumulation, and work done by a force over a distance. In physics problems, particularly those involving variable forces such as gravitational force, integration provides the method to calculate total work done when the force varies over the distance moved.

To find the work done \( W \) when moving an object within a gravitational field, we use the integral of the force over the path taken:

• \( W = \int_{R}^{R+h} F(r) dr \)

This involves integrating the force formula, such as \( \frac{mgR^2}{r^2} \), over the desired limits from the initial position \( R \) to the final position \( R+h \). Integration allows us to sum up the infinitesimal work done at each point along this path:

• Use \( -\frac{1}{r} \) as the integral of \( \frac{1}{r^2} \).
• Evaluate over bounds to find the total work, resulting in an expression like \( \frac{mgRh}{R+h} \).

Understanding how to apply integration enables students to solve complex physics problems involving non-constant forces.

Physics Problem Solving

Approaching a physics problem requires a structured method to break down complex interactions into manageable calculations. Physics problem solving involves understanding the problem, identifying principles involved, deriving equations, performing calculations, and interpreting results. Let's consider a standard approach:

• **Identify Key Concepts:** Recognize the principles and laws at play, such as gravity in the case of launching a rocket.
• **Translate the Problem:** Convert the physical situation into mathematical expressions using related formulas, as we do using \( F = \frac{GmM}{r^2} \) for gravity.
• **Perform Calculations:** Use mathematical techniques such as integration to compute quantities of interest, such as work done, carefully managing constants and limits.
• **Analyze Results:** Interpret the solution in the context of the problem, checking plausibility and consistency with physical laws.

Physics problems often blend various concepts; hence, it’s crucial to practice integrating them effectively. This approach builds the capability to tackle a wide range of scenarios beyond textbook examples.