Question

The energy height, \(H,\) of an aircraft of mass \(m\) at altitude \(h\) and with speed \(v\) is defined as its total energy (with the zero of the potential energy taken at ground level) divided by its weight. Thus, the energy height is a quantity with units of length. a) Derive an expression for the energy height, \(H\), in terms of the quantities \(m, h,\) and \(v\) b) A Boeing 747 jet with mass \(3.5 \cdot 10^{5} \mathrm{~kg}\) is cruising in level flight at \(250.0 \mathrm{~m} / \mathrm{s}\) at an altitude of \(10.0 \mathrm{~km} .\) Calculate the value of its energy height. Note: The energy height is the maximum altitude an aircraft can reach by "zooming" (pulling into a vertical climb without changing the engine thrust). This maneuver is not recommended for a \(747,\) however.

Short answer

Answer: The energy height of the Boeing 747 jet is approximately 13,181 meters.

Step-by-step solution

Find the expression for the potential energy, U

The potential energy, U, is the energy an object has due to its height above the ground. The formula for potential energy is given by: \(U = mgh\) Where m is the mass, g is gravitational acceleration (approximately \(9.81\,\text{m/s}^2\)), and h is the height above the ground.

Find the expression for the kinetic energy, K

The kinetic energy, K, is the energy an object has due to its motion. The formula for kinetic energy is given by: \(K = \frac{1}{2}mv^2\) Where m is the mass and v is the speed of the object.

Derive an expression for the energy height, H

The energy height, H, is the total energy of the aircraft divided by its weight. The total energy is the sum of the kinetic energy and potential energy: \(H = \frac{K + U}{mg}\) Now, replace the expressions for K and U we derived in steps 1 and 2: \(H = \frac{\frac{1}{2}mv^2 + mgh}{mg}\) Simplify the expression: \(H = \frac{v^2}{2g} + h\) This is the expression for the energy height in terms of m, h, and v.

Calculate the energy height for the Boeing 747

We are given mass (\(m = 3.5\cdot 10^5\,\text{kg}\)), speed (\(v = 250.0\,\text{m/s}\)), and altitude (\(h = 10,000\,\text{m}\)). We can now use our expression for the energy height to calculate H: \(H = \frac{v^2}{2g} + h\) Plug in the values: \(H = \frac{(250.0\,\text{m/s})^2}{2(9.81\,\text{m/s}^2)} + 10,000\,\text{m}\) \(H = \frac{62500\,\text{m}^2/\text{s}^2}{19.62\,\text{m/s}^2} + 10,000\,\text{m}\) \(H ≈ 3181\,\text{m} + 10,000\,\text{m}\) \(H ≈ 13181\,\text{m}\) The energy height of the Boeing 747 jet is approximately 13181 meters.

Key concepts

Potential Energy

When studying the energy of an object positioned within a gravitational field, we often refer to potential energy. This is energy stored by virtue of an object's position above a reference point, generally the Earth's surface. For an object of mass m at a height h, the potential energy (U) is calculated using the equation:

[U = mgh]

Here, g represents the gravitational acceleration and is approximately equal to 9.81 m/s² on Earth's surface. Higher altitude not only increases the potential energy due to a greater value of h but also may slightly decrease g as it varies with distance from the Earth's center. For most calculations at or near Earth’s surface, g can be considered constant.

• m (mass) - Typically measured in kilograms (kg)
• g (gravitational acceleration) - A value that, despite mild variation, remains relatively stable near Earth's surface
• h (height) - The vertical displacement from the reference level, measured in meters (m)

When we speak of energy height in the context of an aircraft, we're applying the concept of potential energy to determine how high the aircraft can potentially