Escape Velocity

  • Thread starter MRbrs
  • Start date
  • #1
4
0

Main Question or Discussion Point

Hi guys!

I want to ask something about escape velocity.

I know the definition of EV "escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero"(wikipedia).

Can we approach the problem by equalling the attractive force to the centrifugal force. I tried to find the the V in this way but I found the V [tex]\sqrt{2}[/tex] times smaller. What is wrong with it?

mv^2/r = GmM/r^2

V=[tex]\sqrt{GM/r}[/tex]

sorry about my poor English.
 

Answers and Replies

  • #2
rcgldr
Homework Helper
8,670
506
When velocity is greater than the centripetal force, it just means that the path will not be circular, but it still can be an elliptical orbit. You need sufficient energy for the path to be a parabola or hyperbola.
 
  • #3
4
0
I think when we increase the velocity of the object, centrifugal force becomes greater then attractive force so it escapes. However my equation is still false, why?
 
  • #4
Janus
Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,473
1,178
I think when we increase the velocity of the object, centrifugal force becomes greater then attractive force so it escapes. However my equation is still false, why?
The last post explained why. An increase in velocity will only increase in an elliptical orbit unless it is large enough.

Look at it this way: The object increases speed, so it begins to climb away. However, as it climbs it loses speed in exchange for gravitational potential. Eventually, it reaches a point where it loses enough speed that gravity starts to win the battle again and it begins to fall back. It gains speed at it falls, eventually returning to the point where it started and repeats the cycle again.

The only way that the object can completely escape is for the initial velocity increase to be large enough that the increase in altitude and corresponding decrease in gravity strength keeps ahead of the loss of velocity. This happens when:

[tex]V = \sqrt{\frac{2GM}{r}}[/tex]
 
  • #5
4
0
Thanks rcgldr and Janus for correcting me.

Velocity and force sometimes confuse me. I approached the concept of escape velocity as if the escape force. I guess the only way to determine the velocity is using the kinetic and the potential energies. So thanks again.
 
  • #6
K^2
Science Advisor
2,469
28
What you stumbled on with that radical-2 is Virial Theorem. It states that in any central potential for any closed orbit the average kinetic energy is minus a half of the average potential energy. In order for object to escape, kinetic energy must be equal to minus the potential energy.

In circular orbit, kinetic energy is always the same, so in order to double it, you must increase velocity by square root of 2. So that's exactly the factor by which escape velocity is different from velocity when centrifugal and gravitational forces are balanced.
 
  • #7
4
0
what you stumbled on with that radical-2 is virial theorem. It states that in any central potential for any closed orbit the average kinetic energy is minus a half of the average potential energy. In order for object to escape, kinetic energy must be equal to minus the potential energy.

In circular orbit, kinetic energy is always the same, so in order to double it, you must increase velocity by square root of 2. So that's exactly the factor by which escape velocity is different from velocity when centrifugal and gravitational forces are balanced.
Ok. I first write the equation of forces

mv^2/r = GmM/r^2

then multiplying both sides by r and dividing by 2

1/2 mv^2 = GmM/2r (The equation which you told)

V = [tex]\sqrt{GM/r}[/tex]

this v is true when the objects are balanced. However to find the escape velocity I will have to balance the kinetic energy with potential energy. So I must multiply the velocity by sqrt of 2. Now, everything is clear. Thanks a lot to everyone.
 

Related Threads for: Escape Velocity

  • Last Post
Replies
6
Views
935
  • Last Post
Replies
4
Views
852
  • Last Post
Replies
16
Views
7K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
4
Views
584
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
2
Views
954
Top