Escape Velocity, with varying acceleration as a(y)

Click For Summary

Homework Help Overview

The discussion revolves around determining the minimum initial velocity required for a projectile to escape Earth's gravitational influence, considering acceleration as a function of position. The problem involves concepts from classical mechanics, specifically gravitational acceleration and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between acceleration and velocity, with some suggesting the use of the chain rule to express acceleration in terms of velocity and position. Others mention the potential use of conservation of energy as a method to approach the problem.

Discussion Status

There are multiple lines of reasoning being explored, including integration techniques and energy conservation principles. Some participants have provided guidance on using specific mathematical relationships, while others are questioning the validity of assumptions made regarding constants in their equations.

Contextual Notes

Participants are grappling with the implications of varying gravitational acceleration as the projectile moves away from Earth, which adds complexity to their calculations. There is also a focus on ensuring that the conditions for escape velocity are clearly defined, particularly the requirement that velocity approaches zero as the height approaches infinity.

Chris T.
Messages
4
Reaction score
0

Homework Statement


Find: The minimum initial velocity at which a projectile should be shot vertically from the surface of the Earth so that it does not fall back to Earth. Note: This requires that v=0 as y approaches infinity.

Homework Equations


a = -g[(R^2)/((R+y)^2)]
g = 9.81 m/s^2
R = 6356 km (6.356E6 m)

The Attempt at a Solution


My thought process is to substitute dv/dt for a and integrate to get to velocity, but it seems like I want to find velocity as a function of position? I am stuck, it is the acceleration as a function of position that is confusing me.
 
Physics news on Phys.org
acceleration due to gravity is a function of position
 
Chris T. said:

Homework Statement


Find: The minimum initial velocity at which a projectile should be shot vertically from the surface of the Earth so that it does not fall back to Earth. Note: This requires that v=0 as y approaches infinity.

Homework Equations


a = -g[(R^2)/((R+y)^2)]
g = 9.81 m/s^2
R = 6356 km (6.356E6 m)

The Attempt at a Solution


My thought process is to substitute dv/dt for a and integrate to get to velocity, but it seems like I want to find velocity as a function of position? I am stuck, it is the acceleration as a function of position that is confusing me.
You can use the chain rule to write
$$a = \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v \frac{dv}{dy}.$$
 
Chris T. said:

Homework Statement


Find: The minimum initial velocity at which a projectile should be shot vertically from the surface of the Earth so that it does not fall back to Earth. Note: This requires that v=0 as y approaches infinity.

Homework Equations


a = -g[(R^2)/((R+y)^2)]
g = 9.81 m/s^2
R = 6356 km (6.356E6 m)

The Attempt at a Solution


My thought process is to substitute dv/dt for a and integrate to get to velocity, but it seems like I want to find velocity as a function of position? I am stuck, it is the acceleration as a function of position that is confusing me.

Using conservation of energy is the easiest way to approach this if you know how that works.
 
vela said:
You can use the chain rule to write
$$a = \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v \frac{dv}{dy}.$$
So using the substitution a = vdv/dy and the given equation for acceleration, then separating variables, I should be able to integrate both sides and solve for velocity with respect to y. Let me try that.
 
Dick said:
Using conservation of energy is the easiest way to approach this if you know how that works.
After integration, I find v^2 = 2g(R^2)/(R+y) + C. Is it valid to say that C is zero because v is zero as y goes to infinity?
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K