Re: escape velocity and altitude
- From: "Greg Neill" <gneillREM@xxxxxxxxxxxxxxxxxx>
- Date: Thu, 31 Mar 2005 08:39:18 -0500
"Barry Schwarz" <schwarzb@xxxxxxxxx> wrote in message
news:ivck419ndkq8589ktsl1m0bbvn7po1tv1v@xxxxxxxxxx
> On Tue, 29 Mar 2005 08:24:09 -0500, "Greg Neill"
> <gneillREM@xxxxxxxxxxxxxxxxxx> wrote:
>
[snip]
> >
> >Actually, it doesn't have to be radially directed. So long as the
> >magnitude of the kinetic energy is greater than that of the
> >potential energy at any given point, you're at or above escape
>
> I don't think so. What about being directed radially inward?
It doesn't change the math, but obviously real-world conditions
like the body's paths intersecting should be taken into account
for practical applications.
>
> If you launch at the traditional "escape speed" but directed
> tangentially, several factors contribute to making this value
> insufficient.
Nope. Check the math.
>
> As you travel, your direction is constantly being adjusted
> "downwards" due to gravity.
>
> However, since the speed is in excess of the orbital speed for 0
> altitude, the direction of the gravitational force being applied to
> the object is constantly changing from its starting direction pointing
> at right angles to the direction of travel to a continuously
> increasing direction opposite the direction of travel. (If you launch
> tangentially at orbit speed, the vector of gravitational force remains
> constantly directed radially inward.)
>
> You are gaining altitude at a rate that is a function of both
> gravitational attraction and radius of curvature. (Consider the ideal
> case of Earth with no atmosphere and no gravity. After traveling 4000
> miles tangentially north from the equator, you are "even" with the
> north pole. But you happen to directly above the 45th parallel and
> your altitude is 4000*sqrt(2)-4000 or only 1600 miles. Throw in the
> downward spiral speed and reducing affects of gravity and things look
> worse.)
No. If you launch at escape speed in any direction, the resulting
orbital trajectory is parabolic, or straight line (degenerate conic)
if radially directed.
>
> >speed. Potential energy depends only upon radial position, and
> >kinetic energy depends only on the speed.
>
> While the statement is true, I don't think it applies. A ball
> launched upward with an initial velocity of 32 feet per second will
> travel 16 ft up in one second and retrace its path down in another. A
> ball launched horizontally on the ground (ignoring friction in both
> cases) will never leave the ground. Yet both started with the same
> kinetic and potential energies.
So? You change the shape of the orbit. The body remains bound if
the total mechanical energy is negative, will have a parabolic (escape)
orbit of it is zero, and hyperbolic escape if positive.
>
> >
> >Also, "directed radially" does not give enough information to specify
> >a unique velocity vector.
>
> From any given point, the radial vector from the center of the primary
> is very well defined. I guess for completeness I should have said
> directed away from the center.
It is not a unique vector, so you can't say, "this is the escape velocity,
(x,y,z)".
[snip]
.
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