Problem 42 - Calculus Early Transcendentals

Question

A rocket burning its onboard fuel while moving through space has velocity \(\mathbf{v}(t)\) and mass \(m(t)\) at time \(t\) . If the exhaust gases escape with velocity \(\mathbf{v}_{\text { relative to the rocket, it can be deduced }}\) from Newton's Second Law of Motion that $$m \frac{d \mathbf{v}}{d t}=\frac{d m}{d t} \mathbf{v}_{e}$$ (a) Show that \(\mathbf{v}(t)=\mathbf{v}(0)-\ln \frac{m(0)}{m(t)} \mathbf{v}_{e}\) . (b) For the rocket to accelerate in a straight line from rest to twice the speed of its own exhaust gases, what fraction of its initial mass would the rocket have to burn as fuel?

Step-by-Step Solution

Answer

(a) \( \mathbf{v}(t)=\mathbf{v}(0)-\ln \frac{m(0)}{m(t)} \mathbf{v}_{e} \). (b) The rocket burns approximately 86.5% of its initial mass.

Step 1: Understand the Relationship

The provided equation \( m \frac{d \mathbf{v}}{dt} = \frac{d m}{dt} \mathbf{v}_{e} \) is derived from Newton's Second Law. It states that the force exerted by the gases escaping from the rocket equals the rate of change of the rocket's momentum.

Step 2: Set Up the Integration

To solve for \( \mathbf{v}(t) \), rearrange the equation as \( \frac{d \mathbf{v}}{dt} = \frac{\mathbf{v}_e}{m} \frac{d m}{dt} \). This allows us to integrate concerning time: \( \int \frac{d \mathbf{v}}{dt} \, dt = \int \frac{\mathbf{v}_e}{m} \frac{d m}{dt} \, dt \).

Step 3: Simplify and Integrate

After simplifying, recognize \( \frac{d \mathbf{v}}{dt} dt = d\mathbf{v} \) and \( \frac{\mathbf{v}_e}{m} dm = - \mathbf{v}_e \frac{1}{m} dm \), leading to the integrals: \( \int d \mathbf{v} = - \mathbf{v}_e \int \frac{1}{m} dm \).

Step 4: Evaluate the Integrals

The left integral becomes \( \mathbf{v}(t) - \mathbf{v}(0) \), and the right one evaluates to \( -\mathbf{v}_e \ln m(t) + \mathbf{v}_e \ln m(0) \). Thus, \( \mathbf{v}(t) = \mathbf{v}(0) - \mathbf{v}_e \ln \frac{m(0)}{m(t)} \).

Step 5: Interpret Part (b)

To solve for the fraction of the initial mass burned, set \( \mathbf{v}(t) = 2 \mathbf{v}_e \) and \( \mathbf{v}(0) = 0 \). This results in \( 2 \mathbf{v}_e = -\mathbf{v}_e \ln \frac{m(0)}{m(t)} \). Solve for \( \ln \frac{m(0)}{m(t)} = -2 \) to get \( \frac{m(t)}{m(0)} = e^{-2} \).

Step 6: Compute the Mass Fraction

The fraction of the initial mass burned is \( 1 - \frac{m(t)}{m(0)} = 1 - e^{-2} \). Calculating gives this fraction as approximately \( 0.865 \).

Key Concepts

Newton's Second Law of Motion

Let's start by discussing Newton's Second Law of Motion, which is crucial to understanding rocket propulsion. Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. In formula terms, it's expressed as:

• \( F = m imes a \)

In the context of a rocket, as the rocket burns fuel, the mass of the rocket changes, and so does its velocity. This dynamic scenario results in the rocket's momentum changing over time.
Because the rocket's exhaust gases exert an opposing force, the momentum of the rocket adjusts accordingly. This is captured by the provided equation:

• \( m \frac{d \mathbf{v}}{dt} = \frac{d m}{dt} \mathbf{v}_{e} \)

Here, the term on the left represents the rate of change of the rocket's momentum. The term on the right depicts the force exerted by the escaping gases, reinforcing the direct application of Newton's Second Law in rocket propulsion scenarios.

Momentum

Momentum is a fundamental concept in physics and is intimately connected to Newton's Second Law. It is a measure of an object's motion, defined as the product of its mass and velocity:

• \( p = m imes v \)

For a rocket, momentum shifts as its mass decreases (due to fuel consumption) and its velocity changes. The rocket equation highlights how the rate of change in momentum is influenced by both the changing mass and the velocity of the escaping exhaust gases.
In a rocket propulsion system, the gases escape at high speed, propelling the rocket forward. When the mass of the rocket is reduced as fuel is burned, the velocity must adjust to conserve momentum. Understanding this interplay helps visualize how rockets achieve thrust.

Velocity

Velocity is more than just speed; it is speed with a specific direction. When discussing rockets, velocity determines not only how fast a rocket moves but also the direction in which it travels. In this exercise, the velocity of the rocket (\( \mathbf{v}(t) \)) changes as fuel is expelled.
The derived equation:

• \( \mathbf{v}(t) = \mathbf{v}(0) - \mathbf{v}_e \ln \frac{m(0)}{m(t)} \)

shows how the velocity at any time \( t \) depends on its initial velocity, the velocity of the escaping gases, and the logarithmic relationship with the changing mass. This equation illustrates how losing mass (burning fuel) allows the vehicle to shift its velocity, reaching speeds faster than the exhaust gases themselves.
Deriving this equation involves recognizing that a change in mass affects velocity, crucial for understanding the mechanics of rocketry.

Integration in Calculus

Calculus plays a vital role in solving problems related to rocket motion, especially through integration. Integration helps in finding accumulated quantities, like total change over time. In this case, we solve for velocity by integrating over the variable mass.
The equation:

• \( m \frac{d \mathbf{v}}{dt} = \frac{d m}{dt} \mathbf{v}_{e} \)

is rearranged and becomes:

• \( \frac{d \mathbf{v}}{dt} = \frac{\mathbf{v}_e}{m} \frac{d m}{dt} \)

From here, integration is performed on both sides:

• \( \int \frac{d \mathbf{v}}{dt} \, dt = - \mathbf{v}_e \int \frac{1}{m} \, dm \)

The left side simplifies to velocity change over time, while the right side evaluates through a logarithmic integral. This process leads to calculating a complete equation for velocity, showing calculus's indispensable role in analyzing variable systems like rockets.
Understanding integration allows students to connect how different rates (like mass and fuel consumption) influence outcomes such as velocity.