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An earth satellite of mass \(700 \mathrm{~kg}\) is launched into a free-flight trajectory about the earth with an initial speed of \(v_{A}=10 \mathrm{~km} / \mathrm{s}\) when the distance from the center of the earth is \(r_{A}=15 \mathrm{Mm}\). If the launch angle at this position is \(\phi_{A}=70^{\circ}\), determine the speed \(v_{B}\) of the satellite and its closest distance \(r_{B}\) from the center of the earth. The earth has a mass \(M_{c}=5.976\left(10^{24}\right) \mathrm{kg}\). Hint: Under these conditions, the satellite is subjected only to the earth's gravitational force, \(F=G M_{e} m_{s} / r^{2}\), Eq. \(13-1 .\) For part of the solution, use the conservation of energy.

Step by step solution

01

Understand Gravitational Force

First, understand the formula for the gravitational force: \(F = G M_{e} m_{s} / r^{2}\) where \(M_{e}\) is the mass of the Earth, \(m_{s}\) is the mass of the satellite, \(r\) is the distance between the center of Earth and the satellite, and \(G\) is the gravitational constant \(= 6.67*10^{-11} \mathrm{~Nm}^{2} / \mathrm{~kg}^{2}\).

02

Apply Conservation of Energy

We then apply the conservation of energy principle. Conservation of energy means that the total amount of energy in a system remains constant. Therefore, we can equate the total energy of the satellite at points A and B. Thus, \(\frac{1}{2}m_{s}v_{A}^{2} - \frac{G M_{e} m_{s}}{r_{A}} = \frac{1}{2}m_{s}v_{B}^{2} - \frac{G M_{e} m_{s}}{r_{B}}\). Note that the negative sign for gravitational potential energy is due to the convention that work done to 'lift' the satellite into space is considered as negative as we are moving away from the 'field'.

03

Determine velocity \(v_{B}\) and distance \(r_{B}\)

Next, we simplify the equation and solve for \(v_{B}\) and \(r_{B}\).

04

Convert Units and Find Values

Finally, substitute the known values and convert the units as needed. Remember that \(v_{A}\) is given in km/s and needs to be converted to m/s. \(r_{A}\) is given in Megametre and also needs to be converted to meters.

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Key Concepts

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

Gravitational Force

Gravitational force plays a significant role in satellite mechanics. It refers to the attractive force between two objects with mass. The equation can be expressed as: \[ F = \frac{G M_e m_s}{r^2} \]. This equation tells you a couple of things:

• The force is directly proportional to the masses of the Earth \(M_e\) and the satellite \(m_s\).
• The force is inversely proportional to the square of the distance \(r\) between them.
• The gravitational constant \(G\) gives a universal value of \(6.67 \times 10^{-11} \text{ Nm}^2 / \text{kg}^2\).

This fundamental law is critical in predicting how the satellite will behave over time, especially since no other significant forces are acting upon it in space. Thus, understanding this formula is crucial for any satellite navigation or trajectory calculation.

Conservation of Energy

In satellite mechanics, the conservation of energy principle is used to relate the initial and final states of a satellite in free-flight. This means that the total energy (kinetic plus potential) remains constant unless external forces do work on the system. For our satellite, we can write:\[ \frac{1}{2}m_s v_A^2 - \frac{G M_e m_s}{r_A} = \frac{1}{2}m_s v_B^2 - \frac{G M_e m_s}{r_B} \].

• The kinetic energy \( \frac{1}{2} m_s v^2 \) depends on the speed of the satellite.
• The gravitational potential energy \( -\frac{G M_e m_s}{r} \) depends on the distance from the Earth.

This equation tells us that energy lost in gaining speed is converted from the potential energy field, and vice versa. By knowing the values at one point (such as speed and distance), you can determine these values at another point.

Orbital Motion

Orbital motion refers to the path that the satellite follows as it moves around the Earth under the influence of gravity. This motion is influenced by the initial speed and angle at which the satellite is launched. It can be elliptical, circular, or parabolic depending on its energy and momentum.

• Elliptical orbit occurs when kinetic and potential energies create a balanced condition.
• Circular motion is when a satellite has just the right speed to stay at a constant altitude.
• When speed is extremely high, the satellite may acquire a parabolic or hyperbolic trajectory, potentially escaping Earth's gravity.

Orbital mechanics uses laws of motion and gravity to describe how and when these orbits change over time. Constant observation and calculation help predict the trajectory to ensure successful satellite missions.

Free-flight Trajectory

A free-flight trajectory is the path followed by a satellite not powered by any propulsion engine. It is generally described as being governed purely by gravitational interactions. Once launched, satellites follow a predictable arc as long as no other forces come into play.

• Launch conditions, such as initial speed and angle, significantly influence the resulting trajectory.
• The trajectory is affected by Earth's gravitational pull, which continuously alters the satellite's velocity and position.

Understanding the free-flight trajectory is key for successful satellite deployment and operation. Calculating these paths helps avoid collisions and ensures satellites maintain their intended orbits, effectively fulfilling their communication, observation, or scientific missions.