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Chapter 13: Problem 125

A satellite is launched at its apogee with an initial velocity \(v_{0}=2500 \mathrm{mi} / \mathrm{h}\) parallel to the surface of the earth. Determine the required altitude (or range of altitudes) above the earth's surface for launching if the free-flight trajectory is to be (a) circular, (b) parabolic, (c) elliptical, with launch at apogee, and (d) hyperbolic. Take \(G=34.4\left(10^{-9}\right)\left(\mathrm{lb} \cdot \mathrm{ft}^{2}\right) / \mathrm{slug}^{2}, \quad M_{e}=409\left(10^{21}\right) \mathrm{slug}, \quad\) the earth's radius \(r_{e}=3960 \mathrm{mi},\) and \(1 \mathrm{mi}=5280 \mathrm{ft}.\)

Short Answer

Expert verified
The altitude ranges for different flight trajectories in miles are (a) 1035.91, (b) 2071.82, (c) within the interval (1035.91, 2071.82) and (d) anything above 1035.91.

Step by step solution

01

Convert initial velocity to feet per second

The initial velocity is given in miles per hour. For the ease of computations, it's essential to convert this value into feet per second. This can be done using the conversion factor \(1 \text{ mi/hr} = 1.467 \text{ ft/sec}\). Hence, \(v_{0} = 2500 \times 1.467 = 3675 \text{ ft/sec}\).
02

Compute the Gravitational Parameter

The Gravitational parameter 渭 can be calculated using the formula \(\mu = G \times M_{e}\), where \(G = 34.4 \times 10^{-9} \text{ lb} \cdot \text{ft}^{2}/ \text{slug}^{2}\) and \(M_{e} = 409 \times 10^{21} \text{ slug}\). Therefore, \(\mu = 34.4 \times 10^{-9} \times 409 \times 10^{21} = 1.40736 \times 10^{14} \text{ lb} \cdot \text{ft}^{3}/ \text{sec}^{2}\).
03

Compute the altitudes for different flight trajectories

The altitude for different flight trajectories can be determined through the respective formulas: (a) For a circular orbit, the altitude is given by \(r = \({\mu \over v_{0}^{2}}\), where \(v_{0} = 3675 \text{ ft/sec}\) and \(\mu = 1.40736 \times 10^{14}\). (b) For a parabolic orbit, the altitude is given by \(r = \({2\mu \over v_{0}^{2}}\). (c) For an elliptical orbit, the altitude will lie in the range \(\mu < r < 2\mu \over v_{0}^{2}\).(d) For a hyperbolic orbit, the altitude will be \(\mu > r\).
04

Convert altitude from feet to miles

Conversion factor is \(1 \text{ mi} = 5280 \text{ ft}\). So, each of the altitude value calculated needs to be converted back into miles by dividing the result by 5280.

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Key Concepts

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

Orbital Mechanics
Orbital mechanics, often referred to as celestial mechanics, is a branch of astrodynamics that deals with the motion of satellites and other celestial bodies as they interact under the influence of gravity. The principles of orbital mechanics are crucial for understanding how to send objects into space and ensure they follow a desired trajectory. In the context of satellite launches, the trajectory is the path the satellite takes after launch, and it can vary widely depending on the initial velocity and altitude at which the satellite is launched.

For instance, if the satellite is launched with just the right velocity and altitude, it can enter a circular orbit, continuously falling towards the Earth but with sufficient tangential velocity to keep missing it. If the initial velocity is higher, the satellite might achieve a parabolic or hyperbolic trajectory, where it either escapes Earth's gravity or extends into a far-reaching elliptical orbit. These trajectories are defined mathematically and can be predicted using the laws of physics that dictate how objects move in space under the force of gravity.
Gravitational Parameter
The gravitational parameter, commonly denoted as \(\mu\), is a constant that simplifies the expression of gravitational forces in orbital mechanics. It is the product of the gravitational constant (\(G\)) and the mass of the celestial body (\(M_e\)), such as Earth in the case of a satellite launch. This value gives us a concise measure of the strength of gravity's pull from a particular body, and it plays a critical role in calculating orbits.

Importance of the Gravitational Parameter

Understanding the gravitational parameter is essential to solving problems related to satellite launch trajectories. It allows us to determine the necessary speeds and altitudes for various orbital shapes. For instance, to achieve a circular orbit, the satellite must be launched at a specific altitude that depends on this parameter and the satellite's initial velocity. By calculating the gravitational parameter as shown in the step-by-step solution, one can infer necessary launch parameters for different kinds of orbits, ranging from circular to hyperbolic.
Convert Units
Converting units is a fundamental skill in physics and engineering, necessary for consistent and accurate calculations. The launch velocity of the satellite was initially given in miles per hour, which is not a standard unit in physics calculations. In the exercise, the initial velocity is converted to feet per second (\(ft/s\)) using a conversion factor, which allows for more straightforward computation when working with other units such as the gravitational parameter expressed in feet, pounds, and seconds.

Steps for Unit Conversion

As looked at in the solution, the unit conversion process typically involves multiplying the original measurement by a conversion factor that is equivalent to one. For example, 1 mile is equal to 5280 feet, so to convert from miles to feet, you multiply by 5280. The same process is used to convert from hours to seconds, taking into account the number of seconds in one hour (3600). Understanding how to convert units correctly is critical in solving physics problems to ensure that all the values are in compatible units for the computations to be valid.

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Most popular questions from this chapter

A motorcyclist in a circus rides his motorcycle within the confines of the hollow sphere. If the coefficient of static friction between the wheels of the motorcycle and the sphere is \(\mu_{s}=0.4,\) determine the minimum speed at which he must travel if he is to ride along the wall when \(\theta=90^{\circ}\) The mass of the motorcycle and rider is \(250 \mathrm{kg}\), and the radius of curvature to the center of gravity is \(\rho=20 \mathrm{ft}\) Neglect the size of the motorcycle for the calculation.

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