Animated Solution for Physics - Laws of Motion: A rocket with a lift-off mass $3.5\times 10^4$ kg is blasted upwards with an initial acceleration of $10{\text{ ms}}^{-2}$. Then, the initial thrust of the blast is

The journey to the stars begins with a single, monumental battle against Earth's gravity. When we look at a rocket sitting on a launchpad, we are looking at a masterclass in classical mechanics waiting to happen. In this problem, we are tasked with finding the initial thrust required to blast a $3.5\times 10^4\text{ kg}$ rocket upwards with an acceleration of $10{\text{ m/s}}^2$.

This is not just a plug-and-chug math problem; it is a profound exploration of Newton's Laws of Motion. Let's break down the physics, step by step, and uncover a fascinating anomaly in the provided solution key.

Visualizing the Forces

Imagine you are standing miles away from the launchpad, feeling the low rumble in your chest as the countdown reaches zero. At the exact moment of lift-off, the rocket becomes a battleground for two massive, opposing forces.

First, we have the relentless, invisible hand of gravity pulling the rocket downwards. This is the rocket's weight, mathematically expressed as $mg$. Given the immense mass of the rocket—$3.5\times 10^4\text{ kg}$, or 35 metric tons—this downward pull is colossal.

Second, we have the violent, fiery expulsion of gases from the rocket's engines pushing it upwards. This is the thrust, which we will call $F_t$.

For the rocket to simply hover in place, the thrust must exactly equal the weight. But we don't want to hover; we want to pierce the sky! To accelerate upwards at $10{\text{ m/s}}^2$, the thrust must not only cancel out gravity but also provide an additional net force to generate that upward acceleration.

The Master Equation

To translate this physical reality into mathematics, we call upon the crown jewel of classical mechanics: Newton's Second Law of Motion.

The law states that the net force acting on an object is equal to its mass multiplied by its acceleration:

Let's apply this to our vertical battlefield. We will define the upward direction as positive. Therefore, the upward thrust $F_t$ is positive, and the downward weight $mg$ is negative. Our net force equation becomes:

This is the master equation for rocket lift-off. We want to find the thrust, so let's isolate $F_t$ by moving the weight term to the other side:

This beautifully simple equation tells a profound story: the engine must work hard enough to accelerate the rocket ($ma$) and fight off gravity ($mg$).

Crunching the Numbers

Now, let's bring in our specific numbers and see what the engines need to output.

We know the mass of the rocket:

We know the desired upward acceleration:

And we know the acceleration due to gravity (which we will approximate as $10{\text{ m/s}}^2$ for simplicity, as is common in competitive exams):

Substituting these values into our isolated thrust equation, we get:

First, we add the accelerations inside the parentheses:

Now, we multiply 20 by 3.5. Since $20\times 3.5=70$, our equation simplifies to:

To express this in standard scientific notation, we shift the decimal point one place to the left and increase the exponent by one:

This is the true, physically accurate thrust required to achieve the lift-off described in the problem.

Correcting the Physics

If you look closely at the provided solution key, you will notice something startling. The book claims the answer is option (a), which is $3.5\times 10^5\text{ N}$.

Let's reverse-engineer their logic to see where they went wrong. The book's solution calculates the thrust simply as:

Do you see the critical error? The solution completely ignores the weight of the rocket ($mg$)!

If a rocket engine only produced $3.5\times 10^5\text{ N}$ of thrust, and the rocket weighed $3.5\times 10^5\text{ N}$ (assuming $g=10$), the net force would be exactly zero. The rocket would just sit on the launchpad, burning fuel and making a lot of noise, but it wouldn't move an inch.

To accelerate upwards at $10{\text{ m/s}}^2$, the engine must produce $3.5\times 10^5\text{ N}$ of thrust just to cancel gravity, plus another $3.5\times 10^5\text{ N}$ of thrust to create the upward acceleration.

Therefore, the physically correct answer is Option (b). As an elite student of physics, you must always trust your free body diagrams over a flawed answer key. Never forget gravity!

Rocket Kinematics

This problem captures a single, frozen moment in time: the exact instant of lift-off. But what happens one second later? Or ten seconds later?

A rocket is essentially a giant fuel tank. As it blasts upwards, it consumes fuel at an astonishing rate. This means the mass $m$ of the rocket is not constant; it is rapidly decreasing.

Let's look back at our acceleration equation, rearranged from Newton's Second Law:

If the engine continues to produce a constant thrust $F_t$, but the mass $m$ in the denominator keeps shrinking, the fraction $\frac{F_t}{m}$ will grow larger and larger. Consequently, the upward acceleration $a$ will continuously increase!

This is why astronauts experience intense G-forces as they leave the atmosphere. The rocket gets lighter, but the engines keep pushing just as hard. This dynamic, ever-changing mass system is the foundation of the Tsiolkovsky rocket equation, a beautiful topic you will encounter as you dive deeper into advanced mechanics.

For now, remember the golden rule of dynamics: always draw your free body diagram, account for every force, and never let a flawed answer key shake your confidence in the laws of physics!