How come escape velocity isn't imaginary?

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Discussion Overview

The discussion revolves around the concept of escape velocity in the context of gravitational potential energy. Participants explore the relationship between kinetic energy and potential energy at the point of escape, questioning how escape velocity can be a real number rather than imaginary based on their equations.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants propose that escape velocity is defined as the velocity where kinetic energy equals the potential energy difference between a point of launch and infinity.
  • Others argue that at escape velocity, the sum of kinetic and potential energy is zero, with potential energy being negative and kinetic energy being positive.
  • A participant expresses confusion regarding the derivation of escape velocity leading to an imaginary number and seeks clarification on this point.
  • There is a reiteration of the relationship between kinetic energy and potential energy, with some participants providing similar equations to express their understanding.

Areas of Agreement / Disagreement

Participants express differing views on the definitions and relationships between kinetic and potential energy at escape velocity, indicating that the discussion remains contested without a clear consensus.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the definitions of kinetic and potential energy, particularly concerning the treatment of negative values in potential energy and the implications for escape velocity.

Jules Winnfield
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Going through several definitions, it appears that escape velocity is equal to the potential energy. That is:$$\frac{1}{2}m v^2=-\frac{G M m}{r}$$but if I solve for velocity, $v$, I get:$$v=\sqrt{-2\frac{G M}{r}}$$So how do I get an escape velocity that isn't imaginary?
 
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It's the velocity such that the kinetic energy at launch is equal to the potential energy difference between infinity and the point of launch. So$$\frac 12mv^2=0-\left(-\frac{GMm}r\right)$$
 
Jules Winnfield said:
Going through several definitions, it appears that escape velocity is equal to the potential energy.

No, at escape velocity the sum of kinetic and potential energy is zero.
 
Ibix said:
It's the velocity such that the kinetic energy at launch is equal to the potential energy difference between infinity and the point of launch. So$$\frac 12mv^2=0-\left(-\frac{GMm}r\right)$$
That makes sense. Thank you for the clarification.
 
DrStupid said:
No, at escape velocity the sum of kinetic and potential energy is zero.
And in this sum PE is always negative so KE is always positive.
 

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