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A rocket standing on its launch platform points straight upward. Its engines are activated and eject gas at a rate of \(1500 \mathrm{~kg} / \mathrm{s}\). The molecules are expelled with an average speed of \(50 \mathrm{~km} / \mathrm{s}\). How much mass can the rocket initially have if it is slowly to rise because of the thrust of the engines? The problem provides mass flow and speed, the product of which is equivalent to the time rate-of-change of momentum. That should bring to mind the impulse- momentum relationship, which, of course, is Newton's Second Law. Since the initial motion of the rocket itself is negligible in comparison to the speed of the expelled gas, we can assume the gas is accelerated from rest to a speed of \(50 \mathrm{~km} / \mathrm{s}\). The impulse required to provide this acceleration to a mass \(m\) of gas is $$ F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 $$ from which $$ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{m}{\Delta t} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is \(1500 \mathrm{~kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ An equal but opposite reaction force acts on the rocket, and this is the upward thrust on the rocket. The engines can therefore support a weight of \(75 \mathrm{MN}\), so the maximum mass the rocket could have is $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$

Step by step solution

01

Understand the Problem

The problem asks us to find the maximum initial mass of a rocket, which is determined by the thrust (force) generated from the rocket engines without the rocket rising quickly. We need to use the given details about the mass flow rate (1500 kg/s) and the exhaust speed of the expelled gas (50 km/s) to calculate it.

02

Convert Units

Neither the thrust nor speed is provided in standard SI units. Convert the speed from kilometers per second to meters per second: \(v_f = 50 \times 1000 = 50000\ \text{m/s}\).

03

Calculate Force (Thrust)

The impulse-momentum relationship gives us the thrust (force) that the engine provides. The force can be calculated using the mass flow rate and the speed of expelled gas: \[ F = v_f \times \frac{m}{\Delta t} = 50000\ \text{m/s} \times 1500\ \text{kg/s} = 75,000,000\ \text{N} = 75\ \text{MN} \]

04

Equate Thrust to Weight

Since the upward thrust is equal to the weight of the rocket for it to rise slowly, we set the thrust equal to the weight of the rocket: \[ F = M_{\text{rocket}} \times g \] where \(g = 9.81\ \text{m/s}^2\) is the acceleration due to gravity.

05

Solve for Maximum Rocket Mass

Rearrange the equation to find the mass of the rocket: \[ M_{\text{rocket}} = \frac{F}{g} = \frac{75 \times 10^6}{9.81} = 7,647,466.43\ \text{kg} \] Rounding this off gives approximately \(7.7 \times 10^6\ \text{kg}\) as the maximum mass.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Impulse-Momentum Relationship

The impulse-momentum relationship is a cornerstone of rocket physics. In this context, impulse refers to the change in momentum, which is the product of force and the time over which the force acts. For a rocket, the impulse is created by expelling gas at high speeds from its engines. This causes a momentum change in both the expelled gas and the rocket itself. The formula for impulse is given by:\[I = F \Delta t = \Delta p = m(v_f - v_i)\]where:

• \( I \) is the impulse
• \( F \) is the force exerted
• \( \Delta t \) is the time interval
• \( m \) is the mass
• \( v_f \) and \( v_i \) are the final and initial velocities

The relationship between impulse and momentum helps us understand how the ejection of gas changes the rocket's velocity, crucial for getting the rocket off the ground. Newton's Second Law further helps us relate this impulse to acceleration.

Newton's Second Law

Newton's Second Law states that the force exerted on an object is equal to the mass of that object multiplied by the acceleration it experiences. Formally, it is expressed as \( F = m a \). In the context of rockets, this law helps us determine how the forces from the engines influence the rocket's motion.For our rocket scenario,

• The thrust generated by expelling gas provides the necessary force \( F \).
• The mass \( m \) includes both the structure of the rocket and the fuel.
• The acceleration \( a \) is what allows the rocket to rise.

Substituting into Newton's Second Law:\[F = M_{\text{rocket}} \times g\]where \( g \) is the gravitational acceleration (approx. 9.81 m/s²). Essentially, this allows us to set up an equation where the upward thrust equals the rocket's weight, key to finding the maximum initial mass despite the engines running.

Mass Flow Rate

The concept of mass flow rate is vital in understanding how rockets achieve thrust. This rate reflects the quantity of mass expelled per unit time through the rocket engines, expressed as \( m / \Delta t \). The mass flow rate directly influences the momentum transferred to the expelled gas and hence the thrust generated.For our exercise, the given mass flow rate is:

• \(1500\ \text{kg/s}\)

The greater the mass flow rate, the more effective the generation of thrust, as more matter is expelled in a shorter time, resulting in a significant force produced. The mass flow rate works synergistically with the exhaust speed to build up the necessary momentum for a rocket's movement.

Thrust Calculation

Thrust is the force that propels the rocket upwards, overcoming gravity's pull. Calculating thrust requires the combined data of mass flow rate and the velocity of the expelled gases. The formula to determine the thrust \( F \) is:\[F = v_f \times \frac{m}{\Delta t}\]In this scenario:

• \(v_f = 50000\ \text{m/s}\)
• \(\frac{m}{\Delta t} = 1500\ \text{kg/s}\)
• Thus, \(F = 50000 \times 1500 = 75,000,000\ \text{N} = 75\ \text{MN}\)

The calculation informs us of the rocket's ability to lift off slowly. The significant force illustrates the effectiveness of the engines' design. As long as the thrust equals the gravitational pull on the rocket's mass, it results in gentle upward acceleration, essential for a controlled launch. The ability to calculate thrust is crucial for predicting the maximum mass the rocket can have for a successful launch.