Can you show me algabraically where I messed up on this problem? I need it broken down. I have the question and answer attached, and my work attached as well so you can see what I'm doing wrong. ### Rocket and Gravity: Understanding Distance for Halved Acceleration due to GravityIn this section, we will learn how to calculate the distance a rocket must be from the center of the Earth in order for the acceleration due to gravity to decrease by half.#### Force and Acceleration FormulasStarting with Newton's law of universal gravitation:\[ F = ma = \frac{GMm}{R^2} \]Where:- \(F\) is the gravitational force.- \(m\) is the mass of the object.- \(a\) is the acceleration.- \(G\) is the gravitational constant.- \(M\) is the mass of the Earth.- \(R\) is the distance from the center of the Earth.Next, by isolating the acceleration \(a\), we get:\[ a = \frac{GM}{R^2} \]#### ObjectiveCalculate the distance \( r \) from the center of the Earth where the gravitational acceleration \( a_g \) is half of its usual value at the Earth's surface.#### Calculating the Distance We start with the equation where the acceleration due to gravity is halved:\[ \frac{1}{2} a_g = \frac{GM_{Earth}}{r^2} \]Knowing that \( a_g \) is approximately \( 9.8 \frac{m}{s^2} \), we can substitute in the formula:\[ \frac{1}{2} \left( 9.8 \frac{m}{s^2} \right) = \left( 6.67 \times 10^{-11} \frac{N \cdot m^2}{kg^2} \right) \frac{5.98 \times 10^{24} kg}{(r)^2} \]Solving for \( r \):\[ r = 9.02 \times 10^6 m \]Thus, the rocket must be approximately **9.02 million meters** from the center of the Earth for the acceleration due to gravity to be reduced by half. ### Summary- **Equations used**: Newton's law of gravitation and the formula for acceleration.- **Gravitational constant \(G\)**: \(6.67 \times 10^{-11} \frac{N \cdot m^2}{kg^2}\).- **Earth's ### Topic: Newton’s Law of Universal Gravitation**Objective:**Calculate the distance \( r \) the rocket must be from the center of the Earth for the acceleration due to gravity to decrease by half.**Equations and Constants:**1. **Newton’s Second Law**:\[F = ma\]2. **Newton’s Law of Gravitation**:\[F = G \frac{M_e m}{r^2}\]where:- \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \))- \( M_e \) is the mass of the Earth (\( 5.98 \times 10^{24} \, \text{kg} \))- \( m \) is the mass of the object (rocket)- \( r \) is the distance from the center of the Earth### Step-by-Step Solution:1. **Given**: - Initial acceleration due to gravity at the Earth's surface, \( a = 9.8 \, \text{m/s}^2 \) - Required acceleration due to gravity, \( a' = \frac{1}{2} \times 9.8 \, \text{m/s}^2 = 4.8 \, \text{m/s}^2 \)2. **Calculation**: We want to find \( r \) such that the gravitational acceleration is halved.\[\frac{1}{2} (9.8 \, \text{m/s}^2) = \left(6.67 \times 10^{-11} \, \frac{\text{Nm}^2}{\text{kg}^2} \right) \left( \frac{5.98 \times 10^{24} \, \text{kg}}{r^2} \right)\]3. Solving for \( r^2 \):\[\left( r^2 \right) \left( 4.8 \, \text{m/s}^2 \right) = \left( 6.67 \times 10^{-11} \, \frac{\text{Nm}^2}{\text{kg}^2} \right) \left(

To find distance r with the given conditions.

F = ma = GMm R² M a = G R² Calculate the distance r the rocket must be from the center of the Earth for the acceleration due to gravity to decrease by half. 1 MEarth = G 298 (9.8) = (6.67 x 10-11 N m² 5.98 x 1024 kg kg2 (1)² r = 9.02 x 10 m

F = ma = GMm R² M a = G R² Calculate the distance r the rocket must be from the center of the Earth for the acceleration due to gravity to decrease by half. 1 MEarth = G 298 (9.8) = (6.67 x 10-11 N m² 5.98 x 1024 kg kg2 (1)² r = 9.02 x 10 m

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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