# Why is horizontal escape velocity less ?

1. Nov 22, 2009

### victoriafello

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 im not sure how to work it out ?

2. Nov 22, 2009

### tiny-tim

Hi victoriafello!
(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.

Is this part of a longer question?

3. Nov 24, 2009

### victoriafello

Hi

No its not part of a longer question thats it, the only thing thats 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

4. Nov 24, 2009

### tiny-tim

D'uh! 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 … ?

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?