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Chapter 15: Problem 113

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 \(M_{e}=5.976\left(10^{24}\right)\) 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.

Short Answer

Expert verified
To determine the speed \(v_B\) of the satellite and its closest distance \(r_B\) from the center of the earth, use the conservation of energy principle. Solve for \(v_B\) using the formula: \(v_B = \sqrt{v_A^2 + 2GM_e(1/r_B - 1/r_A)}\) and for \(r_B\) using \(r_B = r_A/(1+(v_A^2r_A/GM_e))\). Computation will give numerical values needed, upon substituting all the known values.

Step by step solution

01

Understanding and Using the Basic Principle

The satellite orbits under the influence of the gravitational force between it and the Earth. The basic principle used here is the conservation of energy. According to it, the total energy (sum of kinetic and potential energy) of an isolated system remains constant if the only forces doing work are conservative forces.
02

Writing the Expression for Total Mechanical Energy

We need to write expressions for kinetic and potential energy of the satellite at points A and B. Potential energy at a distance \(r\) from the Earth's center is given by the formula \(U = -GM_em_s/r\), where \(G\) is the gravitational constant, \(M_e\) is the Earth's mass, \(m_s\) is the mass of the satellite. Kinetic energy at any point on the path is given by, \(K = 0.5*m_s*v^2\). The total mechanical energy at points A and B remains the same, i.e., \(K_A + U_A = K_B + U_B\).
03

Solving for Speed at Point B

Substituting potential and kinetic energies at points A and B in the energy conservation equation, and rearranging terms, we get an expression for \(v_B\): \(v_B = \sqrt{v_A^2 + 2GM_e(1/r_B - 1/r_A)}\). The direction of the velocity doesn鈥檛 factor in, because kinetic energy depends on the square of the speed and is therefore always positive.
04

Solving for Closest Distance From the Earth鈥檚 Center

The satellite's trajectory is a conic section where the energy is negative, and hence, the path is an ellipse. At point B, the distance is smallest hence, the speed is at a maximum. Given \(v_A\) and \(r_A\), he can solve for \(r_B\) using the formula: \(r_B = r_A/(1+(v_A^2r_A/GM_e))\).

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Key Concepts

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

Conservation of Energy
When discussing the motion of satellites and other celestial bodies, the principle of conservation of energy plays a critical role. This principle states that the total energy of an isolated system, like a satellite in space, remains constant over time if it is only subject to conservative forces, such as gravity.

In the context of our exercise, the satellite is in free flight and only influenced by Earth's gravitational pull, making the force acting on it conservative. The satellite's energy is divided into two components: kinetic energy, which is energy due to motion, and potential energy, which is energy due to its position within Earth's gravitational field. Even as the satellite moves and its velocity and height change, the sum of these energies remains unchanged. This conservation allows us to predict the satellite鈥檚 speed at different points in its orbit by equating the total energy at one point with the total energy at another.
Kinetic and Potential Energy
Kinetic energy, represented as \(K=0.5*m_s*v^2\), is the energy that the satellite possesses due to its velocity. It's dependent on both the mass of the satellite \(m_s\) and the square of its velocity \(v\). As a result, even a small change in speed will significantly affect the satellite鈥檚 kinetic energy.

On the other hand, the potential energy in a gravitational field is given by \(U=-GM_e*m_s/r\), where \(G\) is the universal gravitational constant, \(M_e\) is Earth's mass, and \(r\) is the distance from Earth's center. This formula indicates that potential energy is inversely related to the distance: as the satellite gets closer to Earth, its potential energy decreases (becomes more negative), and vice versa. In the satellite's journey, any increase in kinetic energy will correspond to a decrease in potential energy, and the total energy stays the same due to the conservation of energy.
Satellite Trajectory
The path that a satellite takes around a celestial body, like Earth, is known as its trajectory. This trajectory is dependent on the satellite鈥檚 initial speed and the direction of that speed at the time of launch.

In our case, the satellite is launched with an initial velocity and angle that set it on an elliptical path. This path is a conic section, and in an elliptical orbit, the satellite's distance from Earth varies throughout its journey. The conservation of energy principle can be used to determine the satellite鈥檚 distance at various points in its orbit by looking at the changes in kinetic and potential energy. The closest distance from Earth, or the periapsis in an elliptical orbit, is a position of great interest as the satellite attains its highest speed due to gravitational pull.
Earth's Gravitational Force
The force that determines the movement of a satellite around Earth is the gravitational force, which is a central force meaning it is directed along the line joining the satellite and Earth's center. This force is always attractive and its magnitude is given by the formula \(F=G M_e m_s / r^2\), with \(G\) being the gravitational constant, \(M_e\) Earth's mass, \(m_s\) the satellite鈥檚 mass, and \(r\) the distance between the satellite and Earth's center.

Earth's gravitational force is what keeps satellites in orbit and causes the transformation between kinetic and potential energy. The precise calculation of this force is essential when determining the satellite's trajectory and its speed at different points along that trajectory. Notably, this force is the only one acting on the satellite in space, which simplifies our energy conservation equations considerably and makes it possible to solve for unknown variables such as the satellite's velocity and its closest distance to Earth.

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Most popular questions from this chapter

A tankcar has a mass of \(20 \mathrm{Mg}\) and is freely rolling to the right with a speed of \(0.75 \mathrm{m} / \mathrm{s}\). If it strikes the barrier determine the horizontal impulse needed to stop the car if the spring in the bumper \(B\) has a stiffness (a) \(k \rightarrow \infty\) (bumper is rigid), and (b) \(k=15 \mathrm{kN} / \mathrm{m}\).

Disk \(A\) has a mass of \(250 \mathrm{g}\) and is sliding on a smooth horizontal surface with an initial velocity \(\left(v_{A}\right)_{1}=2 \mathrm{m} / \mathrm{s} .\) It makes a direct collision with disk \(B,\) which has a mass of \(175 \mathrm{g}\) and is originally at rest. If both disks are of the same size and the collision is perfectly elastic \((e=1)\) determine the velocity of each disk just after collision. Show that the kinetic energy of the disks before and after collision is the same.

A snowblower having a scoop \(S\) with a crosssectional area of \(A_{\mathrm{s}}=0.12 \mathrm{m}^{3}\) is pushed into snow with a speed of \(v_{s}=0.5 \mathrm{m} / \mathrm{s}\). The machine discharges the snow through a tube \(T\) that has a cross-sectional area of \(A_{T}=0.03 \mathrm{m}^{2}\) and is directed \(60^{\circ}\) from the horizontal. If the density of snow is \(\rho_{s}=104 \mathrm{kg} / \mathrm{m}^{3},\) determine the horizontal force \(P\) required to push the blower forward, and the resultant frictional force \(F\) of the wheels on the ground, necessary to prevent the blower from moving sideways. The wheels roll freely.

A 20 -lb block slides down a \(30^{\circ}\) inclined plane with an initial velocity of \(2 \mathrm{ft} / \mathrm{s}\). Determine the velocity of the block in 3 s if the coefficient of kinetic friction between the block and the plane is \(\mu_{k}=0.25\).

The 20 -kg crate is lifted by a force of \(F=\left(100+5 t^{2}\right) \mathrm{N},\) where \(t\) is in seconds. Determine how high the crate has moved upward when \(t=3\) s, starting from rest.
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