Escape Velocity = Freefall from Height of One Radius?

• fizixfan
In summary: V2 = (2ar)^0.5, where a = surface acceleration of gravity and r = radius of the planetary body. This is derived from the fact that the integral of 1/r^2 is -1/r and that energy per unit mass, corresponding to the latter, differs from acceleration, corresponding to the former, by a factor of r.
Isn't this whole discussion just a little silly? All it amounts to is that if object B (e.g., a rocket) leaves object A (e.g., earth) with enough velocity to reach a height equal to the radius of the object A, it has almost escaped. In fact, if it's initial velocity had been just a bit higher, it would have escaped! So of course the initial velocity needed to reach height h=r is quite close to Vesc (h=∞).

Repeat your calculations to see what initial velocity is needed to reach h=2r and you'll magically find that it is an even closer approximation to Vesc!

phans said:
Isn't this whole discussion just a little silly? All it amounts to is that if object B (e.g., a rocket) leaves object A (e.g., earth) with enough velocity to reach a height equal to the radius of the object A, it has almost escaped. In fact, if it's initial velocity had been just a bit higher, it would have escaped! So of course the initial velocity needed to reach height h=r is quite close to Vesc (h=∞).
Welcome to the Physics Forums!

But by what measure is 70% of escape velocity "quite close"?

jbriggs444 said:
But by what measure is 70% of escape velocity "quite close"?

I also do not understand why people wish to change the definition of escape velocity and agree with you -perhaps the weak field decreasing as inverse square of distance leads them to such calculations of approximation.
however it is best to calculate the potential at the surface of any planet and equate it to the kinetic energy required to escape .

phans said:
Isn't this whole discussion just a little silly? All it amounts to is that if object B (e.g., a rocket) leaves object A (e.g., earth) with enough velocity to reach a height equal to the radius of the object A, it has almost escaped. In fact, if it's initial velocity had been just a bit higher, it would have escaped! So of course the initial velocity needed to reach height h=r is quite close to Vesc (h=∞).
Actually, once that rocket reaches an altitude of 1r, it will at that moment be at rest with respect to the Earth and find itself ~7.9 km/sec short of the velocity it would need to be traveling in order to continue away from the Earth without falling back.
Repeat your calculations to see what initial velocity is needed to reach h=2r and you'll magically find that it is an even closer approximation to Vesc!
Closer, but still ~6.5 km/sec short at the top of the trajectory.

My comment was based on the original premise (post #1) that the acceleration is constant, in which case, quoting from post #7:

"The V2/Vesc values I calculated for the Sun, Earth, Moon and Mars are in 99.9% agreement. I'm not a rocket scientist, but I think this is more that just luck (?)"

That's what I was referring to. The difference in velocity between a drop from height h and from height ∞ obviously approaches 0 as h approaches ∞, so I think my comment stands, given the original assumptions. I was trying to point out that it wasn't just luck; it would have been even better and more mysterious luck if he had arbitrarily started with h=πr.

However, the same principle holds when more accurate, realistic calculations are employed, except that the distance at which V2 and Vesc begin to agree closely moves further out. I didn't bother to calculate the correct h at which V2/Vesc = 0.99.

As the escape velocity is given by
v2=2gR
where R is the radius of the planet and g is the acceleration of gravity on the surface, you can associate the formula with any uniform accelerated motion with acceleration g and over distance R.
And is an exact match, not an approximation. There is nothing special in imagining that R is height above ground. You can go towards the center of the Earth with constant acceleration g or on a tangent or whatever.

If my car reaches 100 km/h over a distance of 100 m, it will have an average acceleration of 3.8 m/s2.
It will reach the same 100 km/h if I drop it from a height of 100 m on Mercury, assuming g has the constant surface value of about 3.8 m/s2. Does this mean that there is a deep connection between Nissan and the Mercurians? :)

Is this such a big subject to discus for 41 posts (including mine :) )

"Is this such a big subject to discus for 41 posts (including mine :) )"

That's precisely the point I was trying to make in post #36 when I said how silly it was to have so many posts going on and on about details that had nothing to do with the original post that used a simple constant g model.

phans said:
"Is this such a big subject to discus for 41 posts (including mine :) )"

That's precisely the point I was trying to make in post #36 when I said how silly it was to have so many posts going on and on about details that had nothing to do with the original post that used a simple constant g model.
The thread had been dead for 48 hours until you woke it up again saying how silly it was that a discussion that had finished was still going on...

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