Why is horizontal escape velocity less ?

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Homework Help Overview

The discussion revolves around the concept of escape velocity in the context of an asteroid with a specified mass and radius. The original poster presents a scenario where the escape velocity differs based on the direction of launch, specifically comparing vertical and horizontal launches.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between escape velocity and the direction of launch, questioning the implications of rotational effects and the definitions of escape velocity versus orbital velocity.

Discussion Status

There is an ongoing exploration of the definitions and implications of escape velocity and orbital velocity. Some participants express confusion about the relationship between the two and the specific conditions of the problem, while others seek clarification on the nature of the launch and its effects.

Contextual Notes

Participants note that the provided escape velocity values may relate to different scenarios: one for escaping the asteroid's gravitational influence and another for achieving a stable orbit. The lack of information regarding the asteroid's rotation is also highlighted as a potential factor in the discussion.

victoriafello
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Homework Statement



an asteriod with mass 6.6x10^17kg & radius 46km has escape velocity 44ms, but if the projectile is launched horizontally then the escape velocity is only 31ms, show that this is correct


Homework Equations



Vesc = SQRT (2GM/R)
G = 6.67x10^-11


The Attempt at a Solution



the only thing i can think of is that the rotational speed of the asteriod is helping to reduce the escape velocity, but I am not sure how to work it out ?
 
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Hi victoriafello! :smile:
victoriafello said:
an asteriod with mass 6.6x10^17kg & radius 46km has escape velocity 44ms, but if the projectile is launched horizontally then the escape velocity is only 31ms, show that this is correct.

(hmm … 31 is approx √(1/2) times 44)

I think that's rubbish …

escape velocity (ignoring rotation) is the same for any direction, since it's the solution to 1/2 mv2 - GM/r2 = 0 + 0.

And although, yes, the rotation of the asteroid can make a difference, you're not given any information about it. :confused:

Is this part of a longer question?
 
Hi

No its not part of a longer question that's it, the only thing that's maybe wasnt clear is that the 31ms is to get the stone into orbit, does that make any difference ?

im really stuck with this one so any ideas would be great
 
victoriafello said:
… the only thing that's maybe wasnt clear is that the 31ms is to get the stone into orbit, does that make any difference ?

D'uh! :rolleyes: yes of course it does!

"escape velocity" means that at large r the speed is zero.

"into circular orbit" means that at large r the speed is proportional to … ? :smile:

But I'm still not understanding the question …

is it a circular orbit? or just a very eccentric elliptical one, always returning to just miss the asteroid?

and does the vertical launch also go into orbit, or does it just keep going straight up? :confused:
 

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