**Problem Statement:**Consider a satellite around Earth in an elliptic orbit. At a point in its orbit, the radius (in km) from the center of the Earth is varying with time as \( r(t) = 7000 - 0.54 t \) (with \( t \) in seconds). The angular rate is \( \dot{\theta}(t) = 1.02 \times 10^{-3} \) rad/s. **Question:**What is the magnitude of acceleration (in m/s²) at \( t = 0 \)?**Solution Guide:**1. **Identify Known Variables:** - Radius as a function of time: \( r(t) = 7000 - 0.54 t \) km - Angular rate: \( \dot{\theta}(t) = 1.02 \times 10^{-3} \) rad/s - Time: \( t = 0 \)2. **Convert Units if Necessary:** - Since the final acceleration needs to be in m/s², convert \( r(t) \) from km to meters: \( r(t) = (7000 - 0.54 t) \times 1000 \) m3. **Evaluate Radius at \( t = 0 \):** \[ r(0) = 7000 \times 1000 \text{ m} = 7 \times 10^6 \text{ m} \]4. **Compute Radial and Tangential Components of Acceleration:** - Radial acceleration (\( a_r \)): \[ a_r = r \ddot{\theta} + \ddot{r} - r \dot{\theta}^2 \] Since \( \dot{\theta} \) is constant: \[ a_r = r \frac{d^2\theta}{dt^2} + \frac{d^2r}{dt^2} - r \left( 1.02 \times 10^{-3} \right)^2 \\ \] Given that \( \frac{d^2r}{dt^2} = 0 \): \[ a_r = 7 \times 10^6 \times 0 - 7 \times 10^6 \times (1.02

Answered: Image /qna-images/answer/31b5801a-43ba-49f2-9686-8ddb45feabb3.jpg

Consider a satellite around Earth in an elliptic orbit. At a point in its orbit, the radius (in km) from the center of the Earth is varying with time as r(t) = 7000 - 0.54 t (with t in seconds). The angular rate is 0 (t) = 1.02 x 10³ rad/s. What is the magnitude of acceleration (in m/s²) at t = 0.

Consider a satellite around Earth in an elliptic orbit. At a point in its orbit, the radius (in km) from the center of the Earth is varying with time as r(t) = 7000 - 0.54 t (with t in seconds). The angular rate is 0 (t) = 1.02 x 10³ rad/s. What is the magnitude of acceleration (in m/s²) at t = 0.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Concept explainers

Basic Terminology in Mechanics

The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its environment. This branch of physics keeps its roots in Ancient Greece in the form of the scripts of Aristotle and Archimedes.

Question

**Problem Statement:**

Consider a satellite around Earth in an elliptic orbit. At a point in its orbit, the radius (in km) from the center of the Earth is varying with time as \( r(t) = 7000 - 0.54 t \) (with \( t \) in seconds). The angular rate is \( \dot{\theta}(t) = 1.02 \times 10^{-3} \) rad/s.

**Question:**
What is the magnitude of acceleration (in m/s²) at \( t = 0 \)?

**Solution Guide:**

1. **Identify Known Variables:**
- Radius as a function of time: \( r(t) = 7000 - 0.54 t \) km
- Angular rate: \( \dot{\theta}(t) = 1.02 \times 10^{-3} \) rad/s
- Time: \( t = 0 \)

2. **Convert Units if Necessary:**
- Since the final acceleration needs to be in m/s², convert \( r(t) \) from km to meters: \( r(t) = (7000 - 0.54 t) \times 1000 \) m

3. **Evaluate Radius at \( t = 0 \):**
\[
r(0) = 7000 \times 1000 \text{ m} = 7 \times 10^6 \text{ m}
\]

4. **Compute Radial and Tangential Components of Acceleration:**
- Radial acceleration (\( a_r \)):
\[
a_r = r \ddot{\theta} + \ddot{r} - r \dot{\theta}^2
\]
Since \( \dot{\theta} \) is constant:
\[
a_r = r \frac{d^2\theta}{dt^2} + \frac{d^2r}{dt^2} - r \left( 1.02 \times 10^{-3} \right)^2 \\
\]
Given that \( \frac{d^2r}{dt^2} = 0 \):
\[
a_r = 7 \times 10^6 \times 0 - 7 \times 10^6 \times (1.02

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

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.

Similar questions

Requlred Informatlon NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A radar system is used to track a new experimental space launch vehicle. Early in the vehicle's flight trajectory, the azimuth angle B Is increasing with the constant rate dßidt= 20/s. The elevation angle y is increasing at the rate dyidt= 40/s, and this rate is increasing at 5/s2. Know that the distance between O and Pis 2 m and that at this instant ß = 0* and y = 30". B dp/dt Determine the velocity and acceleration of point P. The velocity of point Pis (- 0.697 1.21 OJ- 0.603 k) m/s. The acceleration of point P is (- 0.082 OI- +0.539 k) m/s2. 0.34
topic: Space Coordinates & Relative MotionI need this ASAP within 1 hour please
1a) Plot the position in the x-direction as a function of time for both birds on the same co-ordinatesystem (the plots do not have to be exact).b) Write down the equations of motion in x and y directions for bird 1 and 2. c)Using ?⃗ଵ(?) and ?⃗ଶ(?) to represent the position vector of birds 1 and 2, respectively, write downthe components of the position vector for both birds in parenthesis format.d) Write an equation using the tangent of the position vectors that describes when the huntershould release the arrow.e) Solve this equation for the time at which the hunter should release his arrow PLEASE explain, especially 1c, d, and e
please answer this question below. many thanks, sam
1. The wheel starts rotating from rest, is w=0.1.0², where 1) if its angular velocity W as a function of angular position is in radian, and its radius ris 0.2 m, what is the magnitude of the linear velocity of point Pafter the wheel has turned 1.5 revolution(s) from rest. Please pay attention: the numbers may change since they are randomized. Your answer must include 2 places after the decimal point and proper unit. Your Answer: Answer P units
mecha
The graph represents the straight line motion of a car. How far does the car travel between t = 2 seconds and t = 5 seconds? v(mk) 12 4 m 12 m 24 m 36 m 60 m
Please draw the vector diagram and explain the process to solving this question (relative velocity)
The graph in the figure shows the position of an object as a function of time. The letters H-L represent particular moments of time. n H JK Time At which moments shown (H, I, etc.) is the speed of the object (a) the greatest? (b) the smallest? Position O (a) J, (b) L O (a) I, (b) L none of these (a) J. (b) I O (a) I, (b) K O (a) H, (b) L
I need the answer as soon as possible
A mission for disaster monitoring is launched in a circular Sun-Synchronous with semi-major axis a = : 7027.46558 km. a. Compute the orbit period. Also determind how many revolutions the spacecraft completes in a repeat cycle of 19 days? [15 points] b. Compute the distance in km between two consecutive tracks at the equator and show that the minimum distance between tracks in one repeat cycle is 143.125 km. [15 points] c. The Earth orbit angular velocity around the Sun is 0.9860/day. Compute the inclination required for this orbit to be Sun-synchronous, considering that the secular variation of the right ascension of ascending node due to Earth's oblateness is given by 3nJ₂R² 2a² (1 — e²)² in which n is the mean motion of the orbit. Show that the orbit is retrograde. [20 points] d. For this spacecraft, compute the minimum field of view of an onboard camera that enables the global coverage of the Earth. In your calculations you can assume a flat Earth and the camera always pointing in…
Each of four particles move along the x-axis. Their positions (in meters) as functions of time (in seconds) are given as follows: Particle 1: (t) = 3.5 - 2.7³ Particle 2: (t) = 3.5+ 2.7+³ Particle 3: x (t) = 3.5 - 2.7² Particle 4: x (t) = 3.5-3.4t - 2.7t² Which of these particles have constant acceleration? O Only 1 and 2 Only 3 and 4 None of them Only 2 and 3 O All of them