Homework Help is Here – Start Your Trial Now!
Math
Calculus
7. [2/3 Points] DETAILS fre PREVIOUS ANSWERS LARCALCET6 5.1.060. The minimum velocity required for an object to escape Earth's gravitational pull is obtained from the solution of the equation v dv = - GM / / / dy where v is the velocity of the object projected from Earth, y is the distance from the center of Earth, G is the gravitational constant, and M is the mass of Earth. Show that v and y are related by the equation v² = v₁² + 2GM (²-2) where vo is the initial velocity of the object and R is the radius of Earth. To show this, we first evaluate each side [ v dv = = GM] =2 dy ²GM+C When y = R, v = Vo. Using this, we can substitute and solve for the constant of integration, C. v² = GM + C c = ²v₁²- GM R the given equation. Finally, we substitute for C in the original solved equation show how v and y are related by the given equation. GM R 1/v² = GM + (-1/v₁²- GM 22GM+v0²- 2R v² = V₁² + 2GM (= - 2/2 ) Submitted X MY NO

7. [2/3 Points] DETAILS fre PREVIOUS ANSWERS LARCALCET6 5.1.060. The minimum velocity required for an object to escape Earth's gravitational pull is obtained from the solution of the equation v dv = - GM / / / dy where v is the velocity of the object projected from Earth, y is the distance from the center of Earth, G is the gravitational constant, and M is the mass of Earth. Show that v and y are related by the equation v² = v₁² + 2GM (²-2) where vo is the initial velocity of the object and R is the radius of Earth. To show this, we first evaluate each side [ v dv = = GM] =2 dy ²GM+C When y = R, v = Vo. Using this, we can substitute and solve for the constant of integration, C. v² = GM + C c = ²v₁²- GM R the given equation. Finally, we substitute for C in the original solved equation show how v and y are related by the given equation. GM R 1/v² = GM + (-1/v₁²- GM 22GM+v0²- 2R v² = V₁² + 2GM (= - 2/2 ) Submitted X MY NO

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
See similar textbooks
icon
Related questions
Question

What's the variable that will make the proof right then?

7. [2/3 Points] DETAILS The minimum velocity required for an object to escape Earth's gravitational pull is obtained from the solution of the equation [v dv=-GM [ + 2 dy v² = v₁² + 2GM ¹ where v is the velocity of the object projected from Earth, y is the distance from the center of Earth, G is the gravitational constant, and M is the mass of Earth. Show that v and y are related by the equation - 2GM (1/2 - 1 1/2 ) where vo is the initial velocity of the object and R is the radius of Earth. To show this, we first evaluate each side of the given equation. [ v dv = - GM/ dy 2 GM y When y = R, v = Vo. Using this, we can substitute and solve for the constant of integration, C. ܒܟܢ 12v0²: = = GM R C = 12/20²- PREVIOUS ANSWERS LARCALCET6 5.1.060. 12/2v² = v2 = GM y Finally, we substitute for C in the original solved equation to show how v and y are related by the given equation. GM R + C 2GM y + C + GM R + (1/2 v0² - vo²- 2 Vo GM 2.R v² = v₁² + 2GM ( 17 - 12/2) R Submitted X MY NOTES ASK YOUR TEACHER
Transcribed Image Text:7. [2/3 Points] DETAILS The minimum velocity required for an object to escape Earth's gravitational pull is obtained from the solution of the equation [v dv=-GM [ + 2 dy v² = v₁² + 2GM ¹ where v is the velocity of the object projected from Earth, y is the distance from the center of Earth, G is the gravitational constant, and M is the mass of Earth. Show that v and y are related by the equation - 2GM (1/2 - 1 1/2 ) where vo is the initial velocity of the object and R is the radius of Earth. To show this, we first evaluate each side of the given equation. [ v dv = - GM/ dy 2 GM y When y = R, v = Vo. Using this, we can substitute and solve for the constant of integration, C. ܒܟܢ 12v0²: = = GM R C = 12/20²- PREVIOUS ANSWERS LARCALCET6 5.1.060. 12/2v² = v2 = GM y Finally, we substitute for C in the original solved equation to show how v and y are related by the given equation. GM R + C 2GM y + C + GM R + (1/2 v0² - vo²- 2 Vo GM 2.R v² = v₁² + 2GM ( 17 - 12/2) R Submitted X MY NOTES ASK YOUR TEACHER
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS