Question

A rocket of total mass \(1.11 \times 10^{5} \mathrm{~kg}\), of which \(8.70 \times 10^{4} \mathrm{~kg}\) is fuel, is to be launched vertically. The fuel will be burned at the constant rate of \(820 \mathrm{~kg} / \mathrm{s}\). Relative to the rocket, what is the minimum exhaust speed that allows liftoff at launch?

Short answer

To calculate the minimum exhaust speed, the values must be substituted into the formula \(v_{ex} = \frac{M*g}{\dot{m}}\). Upon calculation, the minimum exhaust speed that allows the rocket to lift off can be found and this is the answer to the problem.

Step-by-step solution

List known quantities

The total mass of the rocket \(M\) is \(1.11 \times 10^{5}\) kg. The mass of the fuel \(m_f\) is \(8.7 \times 10^{4}\) kg. The fuel burn rate \(\dot{m}\) is \(820\) kg/s.

Determine the thrust needed for liftoff

The rocket will lift off when the thrust \(F_T\) provided by the exhaust gases equals the weight of the rocket. The weight of the rocket \(F_g\) is the product of its total mass \(M\) and the gravitational acceleration \(g\) approximately \(9.8 \) m/s², hence \(F_g = M*g\). Solving for \(F_T\) gives \(F_T = M*g\).

Determine the exhaust speed

The thrust \(F_T\) is also given by the rate of change of momentum of the exhaust gases, which can be written as \(F_T = \dot{m} v_{ex}\), where \(v_{ex}\) is the exhaust speed. Here \(\dot{m}\) is the mass flow rate, its negative indicates the mass of the rocket is decreasing. Therefore, solving for \(v_{ex}\) gives \(v_{ex} = \frac{F_T}{\dot{m}}\).

Substitute the values

Substituting the given values into the equation, we get \(v_{ex} = \frac{M*g}{\dot{m}}\). Evaluating this expression gives the minimum exhaust speed for the rocket to liftoff.

Key concepts

Thrust and Liftoff

Understanding how a rocket takes off requires a grasp of the concept of thrust. Thrust is the force that moves a rocket through the air and into space. How is it generated? Well, it all comes down to Newton's third law of motion: for every action, there is an equal and opposite reaction.

To achieve liftoff, the thrust must be greater than the weight of the rocket. The weight is the force due to gravity pulling the rocket towards the Earth's center. You can calculate this weight by multiplying the total mass of the rocket, including fuel, with the gravitational acceleration, roughly estimated to be 9.8 meters per second squared on the surface of the Earth. Simply put, for the rocket to begin its journey into space, the thrust needs to overcome gravity.

For the rocket described in the exercise, the thrust needed for liftoff can be calculated by equating it to the rocket's weight. That equation would look like this:
\[ F_T = M \times g \]
Here, \( F_T \) represents the necessary thrust, \( M \) is the rocket's mass, and \( g \) is the gravitational acceleration. Without sufficient thrust, the rocket won't lift off.

Exhaust Velocity

Exhaust velocity is another pivotal factor in rocketry. It pertains to the speed at which gases are ejected from the rocket's engine. Imagine blowing up a balloon and letting it go. The air rushes out of the balloon and it zips away. In a similar fashion, a rocket propels itself forward.

The exhaust velocity is intimately tied to the thrust as shown by the relation:
\[ F_T = \.m v_{ex} \]
The symbol \( \.m \) signifies the rate at which the rocket's fuel is burned and expelled, while \( v_{ex} \) is the exhaust velocity. A higher exhaust velocity means a more efficient thrust and less fuel consumption to achieve lift off. In essence, the minimum exhaust speed is the speed at which the momentum of the expelled fuel is enough to generate a thrust equal to the rocket's weight.

For the rocket in our example, the required exhaust velocity can be computed by re-arranging the formula: \[ v_{ex} = \frac{F_T}{\.m} \]This equation signifies that the exhaust velocity is directly proportional to the thrust needed and inversely proportional to the mass flow rate.

Momentum Conservation

The principle of momentum conservation is a bedrock of physics, particularly when analyzing the motion of rockets. It implies that within a closed system, the total momentum remains constant if no external forces are acting on it.

In the case of a rocket, the system consists of the rocket itself and the exhaust gases. As the rocket burns fuel, it ejects the exhaust gases at high speed in one direction, while the rocket moves in the opposite direction. This is a perfect example of conservation of linear momentum, which is described by the formula:
\[ F_T = \.m v_{ex} \]
The negative rate of the fuel mass flow rate, \( \.m \), signifies the loss of mass (and hence momentum) due to the burning of fuel. The rocket gains momentum in one direction, and the exhaust gases gain an equal amount of momentum in the opposite direction. The conservation of momentum is what allows the rocket to accelerate and, ultimately, to lift off from the ground. This principle is essential when calculating the dynamics of the rocket as it dictates the relationship between the mass of the rocket, the velocity of the exhaust, and the resulting thrust that propels the rocket skyward.