**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

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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

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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

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