Mind Map: Tsiolkovsky’s Rocket Equation and Its Application

The Rocket Equation Explained

Introduction

In this lesson, we explore the intricate details of rocket propulsion and how it aligns with Newton's laws of motion and the law of conservation of linear momentum. Understanding these concepts is crucial for anyone interested in physics and engineering, particularly in the field of rocket science.

The Basics of Rocket Propulsion

Newton's Third Law in Rocket Propulsion

When a rocket expels exhaust gases at high velocity, Newton's third law of motion comes into play. This law states that for every action, there is an equal and opposite reaction. Thus, as exhaust gases are expelled, the rocket propels in the opposite direction.

Conservation of Linear Momentum

The conservation of linear momentum is pivotal in understanding rocket propulsion. As the mass of the rocket decreases due to fuel burning and conversion into exhaust gases, we need to consider how momentum is conserved within this system.

Detailed Derivation of the Rocket Equation

Initial Conditions and Changes Over Time

At any given time T, let's assume a rocket has a velocity V and mass M. After a small time increment ΔT, the rocket's velocity increases to V + ΔV while its mass decreases by ΔM due to fuel burning, resulting in a new mass of M - ΔM.

Applying Conservation of Momentum

To apply the conservation of momentum, we need to consider the initial momentum MV and the momentum after the time increment. The new momentum is given by the sum of the rocket's momentum and the exhaust gases' momentum. By setting these equal, we derive the fundamental relationship:

MV = (M - ΔM)(V + ΔV) + ΔM (V - U)

Simplifying the Equation

Breakdown of Terms

• The term MV represents the initial momentum of the rocket.
• (M - ΔM)(V + ΔV) represents the momentum of the rocket after the time increment.
• ΔM (V - U) represents the momentum of the exhaust gases, where U is the relative velocity of the exhaust gases to the rocket.

Solving the Equation

Expanding and simplifying the equation, we get:

MV = MV + MΔV - ΔMV - ΔMΔV + ΔM(V - U)

By canceling out common terms and ignoring the small term ΔMΔV, we arrive at:

MΔV = UΔM

Final Form of the Rocket Equation

Instantaneous Changes

Taking the time derivative, we get:

M(dV/dt) = U(dM/dt)

Here, (dV/dt) is the rocket's acceleration, and (dM/dt) is the rate at which the rocket's mass changes due to fuel consumption. This can also be written as:

F = U(dM/dt)

Where F is the thrust force experienced by the rocket.

Tsiolkovsky's Rocket Equation

Rearranging the terms and integrating, we get the famous Tsiolkovsky rocket equation:

V - V₀ = U * ln(M₀/M)

Where:

• V is the final velocity of the rocket.
• V₀ is the initial velocity.
• U is the effective exhaust velocity.
• M₀ is the initial mass.
• M is the final mass.

Why do we take -dm instead of dm

The velocity of the exhaust gases with respect to Earth is calculated as (v - u) based on the principle of relative velocities. Here’s a detailed explanation:

1. Relative Velocity Concept : The relative velocity of the exhaust gases u is given with respect to the rocket. This means that if you were sitting on the rocket, you would observe the exhaust gases moving away from you at speed u

2. Rocket’s Velocity with Respect to Earth : The rocket itself is moving with velocity v with respect to an observer on Earth.

3. Combining Velocities : To find the velocity of the exhaust gases with respect to Earth, we need to consider both the velocity of the rocket and the relative velocity of the exhaust gases. If the rocket is moving forward at velocity v, and the exhaust gases are moving backward relative to the rocket at velocity u, the total velocity of the exhaust gases with respect to Earth will be the rocket’s velocity minus the exhaust gases' relative velocity.

Therefore, the velocity of the exhaust gases with respect to an observer on Earth is v-u

To illustrate this with an example :

- Imagine you are on a train (rocket) moving at 50 km/h (v) relative to the ground (Earth).

- If you throw a ball (exhaust gases) backward at 10 km/h (u) relative to the train, the speed of the ball relative to the ground would be the speed of the train minus the speed at which you threw the ball:

50 km/h - 10 km/h = 40 km/h

Similarly, for the rocket and exhaust gases:

- The rocket’s speed relative to Earth is v

- The exhaust gases' speed relative to the rocket is u

Thus, the exhaust gases' speed relative to Earth is v - u