Problem 45 - Calculus with Concepts in Calculus

Question

The mass of a rocket lifting off from earth is decreasing (due to fuel consumption) at the rate of 40 kilograms per second. How fast is the magnitude \(F\) of the force of gravity decreasing when the rocket is 6400 kilometers from the center of the earth and is rising with a velocity of 100 kilometers per second? (Hint: By Newton's Law of Gravitation, \(F=G M m / r^{2}\), where \(G\) is the universal gravitational constant, \(M\) is the mass of the earth, \(m\) is the mass of the rocket, and \(r\) is the distance between the rocket and the center of the earth.)

Step-by-Step Solution

Answer

The magnitude of gravitational force is decreasing continuously due to both the decreasing mass and increasing distance of the rocket.

Step 1: Formulate the Given Problem Variables

Let's assign the known values from the problem: The rate of change of mass of the rocket is \(\frac{dm}{dt} = -40\) kg/s (since the mass is decreasing, it's negative). The distance \(r\) from the center of the earth to the rocket is 6400 km, which we need in meters for consistent units, so \(r = 6400 \times 10^3\) m. The velocity, \(\frac{dr}{dt} = 100\) km/s, is also in meters per second, so \(\frac{dr}{dt} = 100 \times 10^3\) m/s.

Step 2: Differentiate the Gravitational Force Equation

Start by differentiating Newton's Law of Gravitation with respect to time, \(t\). The force is given by: \[ F = \frac{G M m}{r^2} \]We need to find \(\frac{dF}{dt}\), which involves using the chain rule to account for both changing \(m\) and \(r\).

Step 3: Apply the Chain Rule for Differentiation

Using the chain rule: \[ \frac{dF}{dt} = \frac{d}{dt} \left( \frac{G M m}{r^2} \right) = G M \left( \frac{1}{r^2} \frac{dm}{dt} - \frac{2m}{r^3} \frac{dr}{dt} \right) \]This formula accounts for changes in both mass and distance over time.

Step 4: Substitute Known Values Into the Derivative

Now, substitute the known values into the differentiated formula:\[ \frac{dF}{dt} = G M \left( \frac{1}{(6400 \times 10^3)^2} (-40) - \frac{2m}{(6400 \times 10^3)^3} (100 \times 10^3) \right) \]Notice we need some average or unknown mass \(m\) at this point since one part of the rate of change depends on it.

Step 5: Simplify and Calculating the Change in Force

To complete the solution, use estimated or typical values based on reasonable assumptions (like a reference value for \(m\)), or complete the problem without it as an exact example if no mass is specified:Neglecting specifics, \[ \frac{dF}{dt} = -40 \left( \text{constant} \right) - \text{factor} \times m \]This indicates two negative contributions to decreasing \(F\). Notice this step is critical and should capture relevant computation, subject to exact figures of Earth's and rocket's parameters.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a quantity changes. In the context of the exercise, we use differentiation to determine how the gravitational force acting on a rocket changes as certain variables, like mass and distance, change over time.

Since Newton's Law of Gravitation is given by the formula \( F = \frac{G M m}{r^2} \), where \( G \), \( M \), \( m \), and \( r \) represent the universal gravitational constant, mass of the Earth, mass of the rocket, and distance from Earth, respectively, differentiation allows us to see how \( F \) changes when \( m \) and \( r \) change.

To find \( \frac{dF}{dt} \), the differentiation of the force with respect to time, we recognize that both \( m \) and \( r \) are functions of time. Therefore, differentiation will help us break down and understand these distinct contributions to changes in the force of gravity.

Rate of Change

In mathematics, the rate of change refers to how one quantity varies with respect to another quantity. A common way to represent this is through derivatives. In our problem, the rate of change of various parameters needs examination: the mass of the rocket, the gravitational force, and the distance from Earth's center.

Understanding the rate of change is key to predicting how a system behaves over time. In this exercise, the mass of the rocket is decreasing at \( 40 \) kg/s. This is expressed mathematically as \( \frac{dm}{dt} = -40 \). Similarly, the rocket is moving away from the center of the Earth at \( 100 \) km/s, hence \( \frac{dr}{dt} = 100 \times 10^3 \) m/s when converted to a consistent metric system.

Evaluating how these rates affect the gravitational force \( F \) helps in understanding the system's dynamics. By applying these rates into our differentiated formula, we can calculate \( \frac{dF}{dt} \), the rate at which gravitational force decreases.

Chain Rule

The chain rule is a crucial differentiation technique that allows us to differentiate a function composed of other functions. It is particularly useful when variables are interdependent, as seen in many physics problems where multiple variables change over time.

For the problem at hand, Newton's Law of Gravitation involves both the mass of the rocket \( m \) and the distance \( r \) as factors that vary with time. The chain rule provides a systematic method to compute \( \frac{dF}{dt} \), the derivative of the gravitational force with respect to time.

By using the chain rule, we can express:

• \( \frac{dF}{dt} = \frac{d}{dt} \left( \frac{G M m}{r^2} \right) \)
• This expands to: \( G M \left( \frac{1}{r^2} \frac{dm}{dt} - \frac{2m}{r^3} \frac{dr}{dt} \right) \)

Here, each term accounts for how the rate of change of \( m \) and \( r \) impacts \( F \). Through this rule, we receive a comprehensive view, combining the effects of both the decreasing mass and the changing distance on the gravitational force. It essentially ties together the whole system, providing exact insight into how these dynamics operate in tandem to influence \( F \).