Question

A rocket (initial mass \(m_{\mathrm{o}}\) ) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass \(\lambda m_{\mathrm{o}}\) of its fuel, for how long can it hover? [Hint: Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables \(t\) and \(m .\) Take \(v_{\text {ex }}\) to be constant.] (b) If \(v_{\mathrm{ex}} \approx 3000 \mathrm{m} / \mathrm{s}\) and \(\lambda \approx 10 \%,\) for how long could the rocket hover just above the earth's surface?

Short answer

The rocket can hover for approximately 316 seconds, given the specified parameters.

Step-by-step solution

Understand the Thrust and Gravity Balance

In order for the rocket to hover, the thrust produced by the exhaust must equal the force of gravity acting on the rocket. This balance can be mathematically expressed as \( T = mg \), where \( T \) is the thrust, \( m \) is the instantaneous mass of the rocket, and \( g \) is the acceleration due to gravity.

Express Thrust in Terms of Exhaust Velocity

The thrust \( T \) is related to the rate of mass ejection and the exhaust velocity \( v_{\text{ex}} \) by the equation \( T = \dot{m} v_{\text{ex}} \), where \( \dot{m} \) is the rate of change of mass (mass flow rate). For hovering, this equates to \( \dot{m} v_{\text{ex}} = mg \).

Separate Variables for Integration

Rearrange the equation \( v_{\text{ex}} \frac{dm}{dt} = mg \) to separate variables: \[ \frac{dm}{m} = \frac{g}{v_{\text{ex}}} dt. \] This allows us to integrate both sides. The left side integrates with respect to mass, and the right side integrates with time.

Integrate Both Sides

Integrate the mass and time equation from initial mass \( m_{\mathrm{o}} \) to final mass \( m_{\mathrm{o}} - \lambda m_{\mathrm{o}} \) and from time \( 0 \) to \( t \): \[ \int_{m_{\mathrm{o}}}^{m_{\mathrm{o}} - \lambda m_{\mathrm{o}}} \frac{1}{m} \, dm = \int_0^t \frac{g}{v_{\text{ex}}} \, dt \] which gives \( \ln(m) \) evaluated at the limits for the left side and \( \frac{gt}{v_{\text{ex}}} \) for the right side.

Solve the Integrals

Evaluating the left integral gives: \[ \ln(m_{\mathrm{o}} - \lambda m_{\mathrm{o}}) - \ln(m_{\mathrm{o}}) = \ln\left(\frac{m_{\mathrm{o}} - \lambda m_{\mathrm{o}}}{m_{\mathrm{o}}}\right) \] which equals the integral on the right. So, \[ \ln\left(\frac{m_{\mathrm{o}} - \lambda m_{\mathrm{o}}}{m_{\mathrm{o}}}\right) = -\frac{gt}{v_{\text{ex}}}. \]

Solve for Time of Hovering (a)

Exponentiating both sides to solve for \( t \), we have: \[ \frac{m_{\mathrm{o}} - \lambda m_{\mathrm{o}}}{m_{\mathrm{o}}} = e^{-\frac{gt}{v_{\text{ex}}}}. \] Simplifying gives \( 1 - \lambda = e^{-\frac{gt}{v_{\text{ex}}}} \). Taking natural logs gives: \[ -\ln(1 - \lambda) = \frac{gt}{v_{\text{ex}}}, \] which rearranges to solve for \( t \): \[ t = -\frac{v_{\text{ex}}}{g} \ln(1 - \lambda). \]

Calculate the Hover Time (b)

Substitute given values: \( v_{\text{ex}} = 3000 \text{ m/s} \), \( \lambda = 0.1 \), and standard \( g = 9.81 \text{ m/s}^2 \). The hover time \( t \) is: \[ t = -\frac{3000}{9.81} \ln(1 - 0.1). \] Calculate \( \ln(0.9) \), substitute, and solve for \( t \).

Key concepts

Thrust

Thrust is a fundamental concept in rocket mechanics, crucial for understanding how rockets move and hover. It is akin to the pushing force that keeps the rocket in the air by counteracting gravity. Think of thrust as the force generated by expelling gas backward, which propels the rocket forward according to Newton's third law." For the rocket to hover, its thrust must exactly balance the gravitational pull downwards. This balance is described mathematically by the equation \( T = mg \), where \( T \) is the thrust, \( m \) is the rocket's current mass, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \) near Earth's surface).
It means the thrust exerted is just enough to keep the rocket suspended in place without ascending or descending. Whenever thrust exceeds gravity, a rocket will ascend, and if it's less, the rocket will descend." This concept of balancing forces is critical for hovering and provides insight into controlled rocket flight.

Exhaust Velocity

Exhaust velocity is a key driver in determining a rocket's thrust. It refers to the speed at which exhaust gases leave the rocket's engine. A higher exhaust velocity means that the gases are expelled more quickly, which in turn increases the thrust. The relationship between thrust and exhaust velocity can be captured by the formula \( T = \dot{m} v_{\text{ex}} \), where \( \dot{m} \) is the mass flow rate, or the rate at which mass is expelled, and \( v_{\text{ex}} \) is the exhaust velocity.
In essence, exhaust velocity tells us how effectively a rocket engine converts energy into thrust. For instance, if a rocket has an exhaust velocity of \( 3000 \text{ m/s} \), it means that the gases are ejected at that speed. This velocity is vital because it directly influences the amount of thrust generated. The higher the exhaust velocity, the more effective the propulsion system, allowing the rocket to achieve the necessary balance to hover or ascend efficiently.

Mass Flow Rate

Mass flow rate is an essential component in the mechanics of rocket propulsion, indicating how quickly the rocket's fuel mass is being converted to thrust. It is represented as \( \dot{m} \), which is a measure of how many kilograms of fuel are burned per second. The understanding of mass flow rate is vital to managing a rocket's fuel efficiency and determining how long a rocket can maintain thrust for hovering or propulsion.
Mass flow rate is related to thrust and exhaust velocity by the equation \( T = \dot{m} v_{\text{ex}} \). This relationship implies that even if you have a high exhaust velocity, you still need a substantial mass flow rate to generate enough thrust to balance the gravitational force for hovering. Managing the mass flow rate efficiently allows for prolonging the rocket's operational time given a fixed amount of fuel, such as the scenario where a rocket is limited to burning a proportion \( \lambda \) of its initial fuel mass.

Hovering Time

Hovering time is the duration for which a rocket can remain stationary above the ground, effectively balanced between gravity and the thrust from its engines. Calculating hovering time involves integrating the relationship between thrust, mass, exhaust velocity, and gravitational force. With the mathematical setup \( t = -\frac{v_{\text{ex}}}{g} \ln(1 - \lambda) \), where \( \lambda \) is the fraction of the initial fuel mass that can be used, we can determine this time duration.
This formula shows a direct dependency on factors like exhaust velocity (\( v_{\text{ex}} \)) and the fraction of fuel available (\( \lambda \)). A higher exhaust velocity or a larger fuel allocation \( \lambda \) results in a longer hovering duration. As applied to real-life conditions such as \( v_{\text{ex}} = 3000 \text{ m/s} \) and \( \lambda = 0.1 \), with Earth’s gravity \( g = 9.81 \text{ m/s}^2 \), this helps calculate how a rocket can effectively manage its resources to hover for a significant time just above the Earth's surface.