/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 A \(60-\mathrm{kg}\) satellite t... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(60-\mathrm{kg}\) satellite travels in free flight along an elliptical orbit such that at \(A\), where \(r_{A}=20 \mathrm{Mm}\), it has a speed \(v_{A}=40 \mathrm{Mm} / \mathrm{h}\). What is the speed of the satellite when it reaches point \(B\), where \(r_{B}=80 \mathrm{Mm}\) ? Hint: See Prob. \(\quad 14-82, \quad\) where \(\quad M_{e}=5.976\left(10^{24}\right) \mathrm{kg} \quad\) and \(G=66.73\left(10^{-12}\right) \mathrm{m}^{3} /\left(\mathrm{kg} \cdot \mathrm{s}^{2}\right)\)

Short Answer

Expert verified
To find the speed of the satellite at Point B, first calculate the total mechanical energy at Point A as well as the potential energy at Point B. Then use the preservation of mechanical energy to find the kinetic energy and corresponding speed at Point B.

Step by step solution

01

Initial Potential and Kinetic Energy

Firstly, calculate the initial potential and kinetic energy at Point A. Potential energy at a distance \(r\) from the center of the Earth is given by the formula: \(-\frac{GM_{e} m}{r}\), where \(M_{e}\) is Earth's mass, \(m\) is the satellite's mass, and \(G\) is the gravitational constant. Kinetic energy for a moving object with mass \(m\) and speed \(v\) is given by the formula: \(\frac{1}{2} mv^2\). Plug in the given values \(m = 60\), \(r_A = 20\), \(v_A = 40\), \(M_e = 5.976 × 10^{24}\), and \(G = 66.73 × 10^{-12}\) into the formulas and compute the initial potential and kinetic energy.
02

Total Mechanical Energy

Next, add up the initial potential and kinetic energy to obtain the total mechanical energy of the satellite at Point A. This total energy value remains constant throughout the satellite's orbital path under the assumption of an ideal scenario (no energy losses).
03

Potential Energy at Point B

Proceed to calculate the potential energy of the satellite at Point B using the same formula you used in Step 1. However, in place of \(r_A\), use the given \(r_B = 80\) this time.
04

Kinetic Energy and Speed at Point B

The kinetic energy of the satellite at Point B is obtained by subtracting the potential energy at B from the total mechanical energy determined in Step 2. Then, use the formula for kinetic energy \(\frac{1}{2} mv^2 = KE_B\) to find the speed \(v_B\) at Point B. Rearrange the formula to solve for \(v_B\), so that \(v_B = \sqrt{\frac{2KE_B}{m}}\). Calculate the speed at Point B using this formula.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass of the object and its velocity. The formula for calculating kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \]where \(m\) is the mass of the object and \(v\) is its velocity.Understanding kinetic energy can help students solve problems related to motion and energy transformations. In the context of satellite motion, knowing the velocity of the satellite allows us to compute its kinetic energy at any point along its orbit. This energy is pivotal in analyzing changes in speed as the satellite moves through its elliptical path. To find the velocity at a different point in the orbit, you use the principle of conservation of energy. This principle states that, in a closed system, energy cannot be created or destroyed, only transformed. Therefore, the sum of kinetic and potential energy at one point will equal the sum at another, allowing us to solve for unknown speeds.
Potential Energy
Potential energy in the context of orbital mechanics refers to the energy stored due to an object's position in a gravitational field. For a satellite orbiting Earth, this energy is a result of its distance from the Earth's center. The formula for gravitational potential energy is:\[ U = -\frac{G M_e m}{r} \]where \(G\) is the gravitational constant, \(M_e\) is the Earth's mass, \(m\) is the satellite's mass, and \(r\) is the distance from the Earth's center.The negative sign indicates that gravitational force is attractive. It also signifies that energy is needed to move the satellite further away from Earth, increasing its potential energy. In an elliptical orbit, the potential energy varies as the satellite moves closer or farther from Earth. At the closest point, called perigee, potential energy is at its minimum, while kinetic energy is at its maximum due to higher speed. Conversely, at the farthest point, or apogee, potential energy is maximized, and kinetic energy is minimized because of slower speed.
Gravitational Constant
The gravitational constant, symbolized as \(G\), is a key player in the laws of gravity. It is crucial in calculating gravitational forces between two masses and appears in equations for gravitational potential energy and force. The value of \(G\) is approximately \(66.73 \times 10^{-12} \mathrm{m}^3/\left(\mathrm{kg} \cdot \mathrm{s}^2\right)\). This constant provides the scale factor needed to achieve the correct measurements in equations involving gravitational forces. In orbital mechanics, \(G\) helps bridge the connection between the Earth's mass and the satellite's motion. It plays an integral role in determining how the gravitational pull affects the satellite’s velocity and path.Understanding \(G\) is fundamental, as it allows calculations regarding various gravitational interactions not only in orbital mechanics but also in broader contexts like astrophysics and cosmology.
Elliptical Orbit
An elliptical orbit is one where the path taken by an orbiting body, such as a satellite, forms an ellipse rather than a perfect circle. The shape of this orbit is described by its eccentricity, which measures how much it deviates from being a circle. An eccentricity of 0 defines a circle, and values close to 1 indicate more elongated ellipses. In such orbits, the speed of the satellite varies along different points of the path due to the gravitational pull of Earth altering kinetic and potential energies. As the satellite approaches Earth, it moves faster and gains kinetic energy while losing potential energy. Conversely, as it moves away, it slows down, gaining potential energy. The two primary positions in an elliptical orbit are the perigee, the closest approach to Earth, and the apogee, the farthest point. Tracking these changes helps us understand how gravitational forces and energy conservation dictate satellite behavior. Elliptical orbits are essential for satellites tasked with specific earth observation missions, allowing them to cover vast geographic areas effectively.

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Most popular questions from this chapter

If the mass of the earth is \(M_{e}\), show that the gravitational potential energy of a body of mass \(m\) located a distance \(r\) from the center of the earth is \(V_{g}=-G M_{e} m / r\). Recall that the gravitational force acting between the earth and the body is \(F=G\left(M_{e} m / r^{2}\right)\), Eq. \(13-1\). For the calculation, locate the datum at \(r \rightarrow \infty .\) Also, prove that \(F\) is a conservative force.

Escalator steps move with a constant speed of \(0.6\) \(\mathrm{m} / \mathrm{s}\). If the steps are \(125 \mathrm{~mm}\) high and \(250 \mathrm{~mm}\) in length, determine the power of a motor needed to lift an average mass of \(150 \mathrm{~kg}\) per step. There are 32 steps.

A pan of negligible mass is attached to two identical springs of stiffness \(k=250 \mathrm{~N} / \mathrm{m}\). If a \(10-\mathrm{kg}\) box is dropped from a height of \(0.5 \mathrm{~m}\) above the pan, determine the maximum vertical displacement \(d .\) Initially each spring has a tension of \(50 \mathrm{~N}\).

The sports car has a mass of \(2.3 \mathrm{Mg}\), and while it is traveling at \(28 \mathrm{~m} / \mathrm{s}\) the driver causes it to accelerate at \(5 \mathrm{~m} / \mathrm{s}^{2} .\) If the drag resistance on the car due to the wind is \(F_{D}=\left(0.3 v^{2}\right) \mathrm{N}\), where \(v\) is the velocity in \(\mathrm{m} / \mathrm{s}\), determine the power supplied to the engine at this instant. The engine has a running efficiency of \(\varepsilon=0.68\).

As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the \(10-\mathrm{kg}\) block which rests on the smooth surface and is subjected to a horizontal force of \(6 \mathrm{~N}\). If observer \(A\) is in a fixed frame \(x\), determine the final speed of the block if it has an initial speed of \(5 \mathrm{~m} / \mathrm{s}\) and travels \(10 \mathrm{~m}\), both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer \(B\), attached to the \(x^{\prime}\) axis and moving at a constant velocity of \(2 \mathrm{~m} / \mathrm{s}\) relative to \(A\). Hint: The distance the block travels will first have to be computed for observer \(B\) before applying the principle of work and energy.

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