Explanation Given: Satellite is moving around in an elliptical orbit with eccentricity e = 0.130 . The mass of the earth M e = 5.976 × 10 24 kg . The gravitational constant is G = 66.73 × 10 − 12 m 3 / kg ⋅ s 2 The radius of the earth r e = 6.378 × 10 6 m . Write the value for the Viking Explorer velocity at A . υ A = 15 Mm/h = 15 Mm h ( ( 10 ) 6 m 1 Mm ) ( 1 h 3600 s ) = 4166 .6 m/s Write the expression for eccentricity of the earth’s trajectory. e = C h 2 G M s (I) Here, the eccentricity of the earth’s trajectory is e , constant is C , constant is h , the gravitational constant is G and mass of the earth is M e . Write the expression for the maximum distance from the satellite to the centre of Earth. r A = r 0 2 G M e ( r 0 υ 0 ) − 1 (II) Here, minimum distance of the satellite from the center of earth is r 0 and the initial speed of the satellite υ 0 , constant is C , he gravitational constant is G and mass of the earth is M e . Write the expression for the speed of the satellite at maximum distance. υ A = r p r a υ B (III) Here, the speed of the satellite at maximum distance is υ A , the speed of the satellite at minimum distance is υ B , r p is the orbit minimum distance from the center of earth or perigee and r p is the orbit maximum distance from the center of earth or apogee. Write the expression for the distance. d = r 0 − r e (IV) Here, distance is d , minimum distance of the satellite from the center of earth is r 0 and radius of the earth is r e . Conclusion: Substitute 1 r 0 ( 1 − G M s r 0 υ 0 2 ) for C and r 0 υ 0 for h in Equation (I). e = 1 r 0 ( 1 − G M e r 0 υ 0 2 ) ( r 0 υ 0 ) 2 G M e e = ( 1 r 0 − G M e r 0 ( r 0 υ 0 2 ) ) ( r 0 υ 0 ) 2 G M e e = ( r 0 2 υ 0 2 r 0 ( G M e ) − ( r 0 2 υ 0 2 ) G M e ( r 0 2 υ 0 2 ) G M e ) e = ( r 0 υ 0 2 ( G M e ) − 1 ) e + 1 = r 0 υ 0 2 ( G M e ) υ 0 2 = ( e + 1 ) ( G M e ) r 0 1 ( e + 1 ) = ( G M e ) υ 0 2 r 0 r 0 = e + 1 ( G M e ) υ 0 2 (V) Substitute 0

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 13.7, Problem 127P

To determine

The speed when Satellite arrives at apogee A and the distance from the earth’s surface when Satellite is at A .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13.7, Problem 126P

Chapter 13.7, Problem 128P

Students have asked these similar questions

arch Moving to año Question 5 The head-vs-capacity curves for two centrifugal pumps A and B are shown below: Which of the following is a correct statement at a flow rate of 600 ft3/min? Assuming a pump efficiency of 80%. Head [ft] 50 45. 40 CHE 35. 30 25 20 PR 64°F Cloudy 4

I need help with a MATLAB code. I am trying to implement algorithm 3 and 4 as shown in the image. I am getting some size errors. Can you help me fix the code. clc; clear all; % Define initial conditions and parameters r0 = [1000, 0, 0]; % Initial position in meters v0 = [0, 10, 0]; % Initial velocity in m/s m0 = 1000; % Initial mass in kg z0 = log(m0); % Initial mass logarithm a0 = [0, 0, 1]; % Initial thrust direction in m/s^2 (thrust in z-direction) sigma0 = 0.1; % Initial thrust magnitude divided by mass % Initial state vector x0 = [r0, v0, z0] x0 = [r0, v0, z0]; % Initial control input u0 = [a0, sigma0] u0 = [a0, sigma0]; % Time span for integration t0 = 0; % Initial time tf = 10; % Final time N = 100; % Number of time steps dt = (tf - t0) / N; % Time step size t_span = linspace(t0, tf, N); % Discretized time vector % Solve the system of equations using ode45 [t, Y] = ode45(@(t, Y) EoMwithDiscreteMatrix(t, Y, u0, x0, t0, tf), t_span, x0); % Compute the matrices A_k,…

Q2) Determine the thickness of weld (h) for the figure shown below. when the Su= 410 MPa and factor of safety of 2. COR 50 200 60 F=2000N

Chapter 13 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 13.4 - The 10-kg block is subjected to the forces shown....Ch. 13.4 - The 10-kg block is subjected to the forces shown....Ch. 13.4 - Determine the initial acceleration of the 10-kg...Ch. 13.4 - Prob. 4PP Ch. 13.4 - Prob. 1FP Ch. 13.4 - If motor M exerts a force of F = (10t2 + 100) N on...Ch. 13.4 - A spring of stiffness k = 500 N/m is mounted...Ch. 13.4 - Prob. 5FP Ch. 13.4 - Block B rests upon a smooth surface. If the...Ch. 13.4 - The 6-lb particle is subjected to the action of...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The speed when Satellite arrives at apogee A and the distance from the earth’s surface when Satellite is at A .

Solution Summary: The author determines the speed when Satellite arrives at apogee A and the distance from the earth’s surface.

Concept explainers

Videos

The speed when Satellite arrives at apogee A and the distance from the earth’s surface when Satellite is at A .

Want to see the full answer?

Chapter 13 Solutions