PDA

View Full Version : escape velocity

stitt29
2009-May-29, 10:57 AM
Hi,

I've read that the escape velocity for the Earth is 11000km/sec. I've also read that if you kick a football it is on an orbital path. The reason it doesn't continue on this path is that it hits the body of the Earth as it falls. Does this mean that firing a projectile at 11000km/sec will enable the path to be so wide/high or whatever that it orbits without collission with the Earth. Less than 11000km/sec will mean it eventually hits the Earth; more than 11000km/sec it will orbit in a higher path or larger orbit around the Earth (or more accurately the barycentre of the 2 objects).

grant hutchison
2009-May-29, 11:09 AM
That's 11 km/s, or 11000 m/s. :)
You can get into orbit with a lower velocity. Firing something horizontally at 8 km/s (and neglecting air resistance) will give you an orbit that doesn't ever fall low enough to strike the earth. A little slower than that, and your projectile will always "try" to drop lower than the earth's surface at some point in its orbit, no matter what direction you fire it in.
Faster than 11 km/s, and your projectile will rise constantly and never return to earth. (Provided you don't fire it at the ground in the first place!)
Between 8 and 11 km/s, you'll get a family of elliptical orbits.

Grant Hutchison

saturn4b
2009-May-29, 11:14 AM
Hi,

I've read that the escape velocity for the Earth is 11000km/sec. I've also read that if you kick a football it is on an orbital path. The reason it doesn't continue on this path is that it hits the body of the Earth as it falls. Does this mean that firing a projectile at 11000km/sec will enable the path to be so wide/high or whatever that it orbits without collission with the Earth. Less than 11000km/sec will mean it eventually hits the Earth; more than 11000km/sec it will orbit in a higher path or larger orbit around the Earth (or more accurately the barycentre of the 2 objects).

If you know a trick for getting spacecraft up to 11000km/sec, I think NASA would be very interested. :)

The speed of 11km/sec (approx) is the escape velocity at the height of low earth orbit - the kind of speed you did on Apollo to get clear of the earth and get to the moon. If you are orbiting, say, 30,000km out, then you'd not have to get up to quite as high a speed to leave orbit. Imagine, if you will, heading straight up from the earth at high speed (assuming no atmosphere) - as you ascend, gravity will slow you down more and more. Escape velocity is going fast enough that gravity can't slow you down enough to stop you getting away. Less than that means you will slow down, stop and then fall back.

At a lower velocity, you can sustain orbit. In a way you are always falling to the earth but your movement sideways means you keep missing!

Peter B
2009-May-29, 11:25 AM
Hi,

I've read that the escape velocity for the Earth is 11000km/sec.

G'day Stitt

Not quite that high! It's just over 11 km/sec, or 11,000 metres/sec.

I've also read that if you kick a football it is on an orbital path. The reason it doesn't continue on this path is that it hits the body of the Earth as it falls.

Essentially, yes.

Does this mean that firing a projectile at 11000km/sec will enable the path to be so wide/high or whatever that it orbits without collission with the Earth. Less than 11000km/sec will mean it eventually hits the Earth; more than 11000km/sec it will orbit in a higher path or larger orbit around the Earth (or more accurately the barycentre of the 2 objects).

Not quite.

Firstly, (assuming you aim the projectile above the horizon!) at speeds over 11 km/s, the projectile will depart from the Earth, never to return. Secondly, (if I've got it right) at speeds between about 8 and 11 km/s, the projectile will orbit the Earth. Thirdly, at speeds below 8 km/s, the projectile will return to the Earth.

The 8 km/s speed is the notional orbital velocity at the surface of the Earth, and is lower at higher altitudes. I have a vague memory that escape velocity is equal to orbital velocity times the square root of two, but the sum doesn't seem to work.

I'm sure someone more knowledgeable than me will be able to correct my mistakes.

Edit to add: Woo-hoo! Judging from Grant Hutchison's answers, I think I got it right!

NEOWatcher
2009-May-29, 12:13 PM
...The 8 km/s speed is the notional orbital velocity at the surface of the Earth, and is lower at higher altitudes. I have a vague memory that escape velocity is equal to orbital velocity times the square root of two, but the sum doesn't seem to work...
Yes; but you're working with rounded numbers. Even with just one more decimal of accuracy, the formula works out much better.
7.8km/s and 11.2km/s

stitt29
2009-May-29, 12:21 PM
superb answers. Thanks very much boys. 11km/sec thanks for the correction.
This raises one further question though

Faster than 11 km/s, and your projectile will rise constantly and never return to earth.
Does this mean the orbit will be forever increasing like a spiral?

I've heard it said if the orbital speed of any planet was less it would fall into the Sun If greater it would leave the solar system. From what I can gather any planet would not leave the solar system, if it had a greater orbital speed ,but would rise to a higher orbit. Is this right?

Hornblower
2009-May-29, 12:32 PM
superb answers. Thanks very much boys. 11km/sec thanks for the correction.
This raises one further question though

Does this mean the orbit will be forever increasing like a spiral?Not a spiral, but the outbound portion of a parabola or hyperbola.

I've heard it said if the orbital speed of any planet was less it would fall into the Sun If greater it would leave the solar system. From what I can gather any planet would not leave the solar system, if it had a greater orbital speed ,but would rise to a higher orbit. Is this right?
Suppose we start with a circular orbit. If we slow the object down, it will drop into an elliptical orbit with a perihelion inside the original orbit, and an aphelion at or near the point of the slowdown. We would need to slow it down almost to a dead stop to make the perihelion encroach on the Sun's surface.

If we speed it up by less than (sqrt 2) times the original velocity, it will rise into an elliptical orbit with the aphelion higher than the original orbit and the perihelion at or near the original orbit. If we speed it up by factor of sqrt 2 or more, it will escape.

NEOWatcher
2009-May-29, 12:34 PM
Does this mean the orbit will be forever increasing like a spiral?
No; spirals don't happen in orbital mechanics unless there is some kind of drag (ex. atmospheric, tidal) or other forces that disturb the orbit.

Below escape velocity, the orbit is an ellipse, with the circle being a specific case of that ellipse.
Below orbital velocity, that ellipse crosses the surface of the Earth. That's why it falls to Earth.
At escape velocity, that ellipse opens up to a parabola.

I've heard it said if the orbital speed of any planet was less it would fall into the Sun If greater it would leave the solar system. From what I can gather any planet would not leave the solar system, if it had a greater orbital speed ,but would rise to a higher orbit. Is this right?
Kind of both right if you add the right adjectives. Huge changes for the first part of that phrase.

If the change is one time, the planet will just change to an elliptical orbit. It takes another adjustment to circularize it to a higher or lower orbit.

grant hutchison
2009-May-29, 03:42 PM
No; spirals don't happen in orbital mechanics unless there is some kind of drag ... "A spiral is not a compatible orbit," to quote Robert Heinlein from Starman Jones. :)

Grant Hutchison

mugaliens
2009-May-29, 05:15 PM
Below escape velocity, the orbit is an ellipse, with the circle being a specific case of that ellipse.
Below orbital velocity, that ellipse crosses the surface of the Earth. That's why it falls to Earth.
At escape velocity, that ellipse opens up to a parabola.

And beyond escape velocity (since hitting ev precisely is such a difficult thing, and not very useful) it's a hyperbola... :)

NEOWatcher
2009-May-29, 05:24 PM
And beyond escape velocity (since hitting ev precisely is such a difficult thing, and not very useful) it's a hyperbola... :)

And speaking of, and insistance upon an exact parabola could be a hyperbole. :lol:

cjameshuff
2009-May-29, 07:36 PM
No; spirals don't happen in orbital mechanics unless there is some kind of drag (ex. atmospheric, tidal) or other forces that disturb the orbit.

Well...a diagram will show a trajectory that wraps around the planet to some degree before leaving the plotted area. That could legitimately be called a spiral, even though it never completes a revolution. And if you plot the trajectory using a coordinate system fixed to the rotating Earth, the result will be a spiral.

Either of those might be sources of the misconception that orbits are discrete things that objects will spiral out of at the slightest disturbance. As mentioned by others, though, aside from situations involving drag or constant thrust, or weird relativistic effects around compact objects like black holes, it is not useful to describe objects as spiraling in or out...at best it's misleading, and it's usually just wrong.

To recap what other's have already said: For an object in circular orbit around a single body, a change in the orbital speed of the orbiting object that leaves it less than escape velocity will just shift it into an elliptical orbit that touches the original circular orbit at either closest approach or furthest extent from the central body. The opposite can also be done, circularizing an elliptical orbit. Exactly hitting escape velocity will cause it to leave on a parabolic trajectory, and if greater than escape velocity is reached, it will leave on a hyperbolic trajectory. In either of those cases, its direction of motion will never turn back toward the central body...its distance will increase forever.

NEOWatcher
2009-May-29, 07:53 PM
Well...a diagram will show a trajectory that wraps around the planet to some degree before leaving the plotted area. That could legitimately be called a spiral, even though it never completes a revolution. And if you plot the trajectory using a coordinate system fixed to the rotating Earth, the result will be a spiral.
While I realize you are describing a misconception, I disagree that it could "legitimately" be called a spiral. I could go with "interpreted" as.

The definition (http://dictionary.reference.com/browse/SPIRAL) uses words like "around", "ring" and "circle" which imply 360 degrees. Plus it says a "fixed point" of which rotation puts a wrench in the works.

Amber Robot
2009-May-29, 08:27 PM
While I realize you are describing a misconception, I disagree that it could "legitimately" be called a spiral. I could go with "interpreted" as.

Agreed. All non-decaying (or the opposite) orbits are conic sections. A spiral is not a conic section. So the word 'legitimately' is misplaced here.

a1call
2009-May-29, 09:01 PM
Agreed. All non-decaying (or the opposite) orbits are conic sections. A spiral is not a conic section. So the word 'legitimately' is misplaced here.

The Moon's orbit is non-decaying and it's current increasing distance (http://www.newton.dep.anl.gov/askasci/ast99/ast99639.htm) is probably better described as spiraling outward than a static conic section.

Of course as mentioned by others oceanic tidal drag (http://www.newton.dep.anl.gov/askasci/ast99/ast99657.htm) is the cause.

Amber Robot
2009-May-29, 09:15 PM
The Moon's orbit is non-decaying and it's current increasing distance (http://www.newton.dep.anl.gov/askasci/ast99/ast99639.htm) is probably better described as spiraling outward than a static conic section.

That's why I said "(or the opposite)". I just didn't have at hand a good term for the opposite of non-decaying.

And before some smarty-pants comes in and brings up precession, I'll say "yeah, yeah... sheesh" in advance. :p

cjameshuff
2009-May-29, 09:27 PM
The definition (http://dictionary.reference.com/browse/SPIRAL) uses words like "around", "ring" and "circle" which imply 360 degrees.

Aside from that not being a formal mathematical definition, that's a very weak implication, especially for the first definition which uses "around". Starting from the surface of a planet, a straight line from the origin point to a point at infinity will generally pass through the planet. How do you get from here to there without going around the planet?

There is a non-zero angular velocity for most escape velocities. The definition of spiral says nothing about the relationship of angular to radial velocity, and does not require that angular velocity of a point moving along the curve remain nonzero for a full rotation. It seems perfectly legitimate to consider a parabolic or even a hyperbolic curve originating from a point on and along a tangent to a circle to be spiraling out from it.

Plus it says a "fixed point" of which rotation puts a wrench in the works.

No it doesn't. That the plotted shape is of coordinates in a rotating reference frame is irrelevant to what the plotted shape *is*. Such a plot is of limited use, but it quite definitely is both a plot of an orbital trajectory and a spiral.

NEOWatcher
2009-Jun-01, 12:01 PM
Plus it says a "fixed point" of which rotation puts a wrench in the works.
No it doesn't.
Did you even open the definition I provided?

1. Geometry. a plane curve generated by a point moving around a fixed point while constantly receding from or approaching it.

Now; I am willing to take an open look at many of the things you are saying, but when omissions like this are made, I tend to disregard the arguments.

The definition of spiral says nothing about the relationship of angular to radial velocity, and does not require that angular velocity of a point moving along the curve remain nonzero for a full rotation.
But; it does say "constantly". The recession of a conic section is not constant.

While I am understanding that you may be using "spiral" in the common usage, we are talking about a geometric concept. In that context, it is not a spiral, otherwise, I wouldn't have such a stringent view of it. After all, this is a scientific board.

stitt29
2009-Jun-01, 12:28 PM
another couple of points I want to raise.

Does it matter the angle you point your projectile at i.e. Say I fire a ball at 10degrees above the horizon at escape velocity and another at the same speed, angle 45degrees. will both balls settle into an elliptical orbit of the same length/time period.

Also if something is fired at above ev the ball should escape but it will still be orbiting the Sun so what is it's orbit like. I thought it would still have to orbit the Earth as it will still be pulled by it gravitationally (to some degree).Which means it hasn't escaped really. What I'm thinking is nothing can totally escape another bodies gravity even if it gets out of the solar system. Anyway please put me straight on these points.

chornedsnorkack
2009-Jun-01, 01:32 PM
There is a non-zero angular velocity for most escape velocities.

But not all.

An object may escape at zero angular velocity. In which case, an item thrown directly upwards at exactly the escape speed will rise vertically at an ever slower speed, and slow down to arbitrarily slow speed, but never stop altogether and fall back down.

Escape velocity is completely independent on direction. The same object thrown directly horizontally will follow a parabola and perform just less than 180 degrees orbit around the Earth. Again, it slows down to an arbitrarily slow speed, but never stops, and never completes 180 degrees.

PraedSt
2009-Jun-01, 03:51 PM
Does it matter the angle you point your projectileIn general, yes.

Say I fire a ball at 10degrees above the horizon at escape velocity and another at the same speed, angle 45degrees. will both balls settle into an elliptical orbit of the same length/time period.It might be better if you try and not think that escape velocity is something you want to hit; it's something you want to exceed.

Anyway, if you do hit it exactly, as NEOWatcher said above, your orbit will be a parabola, not an ellipse. Look these up! :)

So, if you fire two balls at two different angles, you will get two different parabolic 'orbits'.

Also if something is fired at above ev the ball should escape but it will still be orbiting the Sun so what is it's orbit like.If the speed is above Earth's escape velocity, but below the Sun's escape velocity, it'll orbit the Sun. It'll be like any other orbit; an ellipse. The 'shape' of the orbit won't be almost circular, like those of our planets, but more 'stretched', similar to those of the comets.

I thought it would still have to orbit the Earth as it will still be pulled by it gravitationally (to some degree).Which means it hasn't escaped really. What I'm thinking is nothing can totally escape another bodies gravity even if it gets out of the solar system. Anyway please put me straight on these points.You're right in a way. You can't totally escape another body's gravity. But what's more important is which body is the dominant 'source' of gravity. At a certain distance form Earth, its gravity becomes so small relative to the gravity of the Sun, that you can discount it. For all intents and purposes, the Sun will be the only body that counts. Similarly, if you escape the Solar System, the influence of the Sun relative to the influence of the Galaxy as a whole will diminish. You'll then be orbiting the Galactic centre, and you can forget about the feeble Sun.

cjameshuff
2009-Jun-01, 08:35 PM
Did you even open the definition I provided?

Yes I did. I read all of them. They all left room for interpretation.

Now; I am willing to take an open look at many of the things you are saying, but when omissions like this are made, I tend to disregard the arguments.

Omissions like what? As I pointed out, "around" does not imply following a full 360 degree path around a point, and the central point need only be fixed in position on the plot...it's not suddenly not-a-spiral because the plot happens to physically relate to a position in a rotating coordinate system.

But; it does say "constantly". The recession of a conic section is not constant.

The exact same curve can be drawn using constant radial velocity, constant angular velocity, constant path velocity, or with none of these being constant. Are you sure it's only a spiral in the first case?

That definition is from a popular dictionary, not a mathematics textbook. It says constantly, but it probably means monotonically, distance from the center always increasing with distance along the path. Even that is not always the case, since there are spiral curves with more than one center where this is only true for portions of the curve and specific centers.

While I am understanding that you may be using "spiral" in the common usage, we are talking about a geometric concept. In that context, it is not a spiral, otherwise, I wouldn't have such a stringent view of it. After all, this is a scientific board.

Er...you're the one using a fuzzy, common language definition from an online English dictionary to support your point.

But not all.

Irrelevant. I'm not arguing that all trajectories or all plots of trajectories are spirals. Escape trajectories just happen to be a case where the word is both technically valid in the general case and has a high risk of being misunderstood.

Back to the actual subject of the thread...

Does it matter the angle you point your projectile at i.e. Say I fire a ball at 10degrees above the horizon at escape velocity and another at the same speed, angle 45degrees. will both balls settle into an elliptical orbit of the same length/time period.

At escape velocity, both will leave the planet, never to return. Neither will enter elliptical orbits, neither trajectory will have a period...if there are no other sources of gravity, they will continually slow, forever approaching zero velocity at infinite distance after an infinite period of time. If they start off with higher than escape velocity, they instead approach a velocity higher than escape velocity.

Also if something is fired at above ev the ball should escape but it will still be orbiting the Sun so what is it's orbit like. I thought it would still have to orbit the Earth as it will still be pulled by it gravitationally (to some degree).Which means it hasn't escaped really. What I'm thinking is nothing can totally escape another bodies gravity even if it gets out of the solar system. Anyway please put me straight on these points.

Escape doesn't mean that a body has left the gravitational influence of another body, it means that it is no longer gravitationally bound to it. At a given location in a gravity well, an object has a certain negative potential energy, equal in magnitude to that gained by that object falling from infinity to that point. If it gains an equal amount of kinetic energy, it now has enough energy to leave that gravity well entirely...it is still affected by that object's gravity, but has enough kinetic energy to escape it.

solomarineris
2009-Jun-01, 10:48 PM
Suppose that I take off vertically with slowest rocket, say 200mph, if my speed is steady, won't I reach the orbit eventually?
If I continue with same speed against Earth's gravity won't I leave the Planet's grip eventually?
I guess what I fail to get is, why escape velocity is relevant to leave the grasp of Earth's gravity.
Can't we get up there (orbit) at our leisure? Like space transporters do in Sci-fi movies?

Nick Theodorakis
2009-Jun-01, 11:00 PM
Suppose that I take off vertically with slowest rocket, say 200mph, if my speed is steady, won't I reach the orbit eventually?
If I continue with same speed against Earth's gravity won't I leave the Planet's grip eventually?
I guess what I fail to get is, why escape velocity is relevant to leave the grasp of Earth's gravity.
Can't we get up there (orbit) at our leisure? Like space transporters do in Sci-fi movies?

You could get to space that way, but if you turned off your rocket you will fall back to earth (or the sune, I guess, if you were far enough away) unless you increased your speed to orbital velocity or greater.

Nick

solomarineris
2009-Jun-01, 11:10 PM
You could get to space that way, but if you turned off your rocket you will fall back to earth (or the sune, I guess, if you were far enough away) unless you increased your speed to orbital velocity or greater.
Nick

Suppose that I have a F/A-22 Raptor, airtight, with 2 machs at my disposal, suppose also I have fairly good fuel reserves, can't I go anywhere I want?
Taking my sweet time?

Nick Theodorakis
2009-Jun-01, 11:12 PM
Anywhere there is enough air to feed into the engine. What's its ceiling, 60,000 ft. or something?

Nick

Amber Robot
2009-Jun-01, 11:18 PM
I guess what I fail to get is, why escape velocity is relevant to leave the grasp of Earth's gravity. Can't we get up there (orbit) at our leisure?

In the context of the current discussion, orbit would not be considered outside the grasp of Earth's gravity.

solomarineris
2009-Jun-01, 11:52 PM
Anywhere there is enough air to feed into the engine. What's its ceiling, 60,000 ft. or something?
Nick
Nick,
I wasn't talking about a conventional aircraft, this hypothetical craft would be a spacecraft with the same powers, it wouldn't have turbofan engines since there's no air up there.
What I am trying to visualize is; Is it possible to escape Earth's grip with minimum speed, if the craft continues to climb, like, in a way Space Shuttle, which has been it's highest orbit just to fix Hubble.

Hornblower
2009-Jun-02, 01:15 AM
Suppose that I take off vertically with slowest rocket, say 200mph, if my speed is steady, won't I reach the orbit eventually?
If I continue with same speed against Earth's gravity won't I leave the Planet's grip eventually?
I guess what I fail to get is, why escape velocity is relevant to leave the grasp of Earth's gravity.
Can't we get up there (orbit) at our leisure? Like space transporters do in Sci-fi movies?That calculated escape velocity is what it takes to escape with no further thrust after reaching that velocity.

If you had the rocket of our dreams that could keep you ascending vertically at 200mph for a long enough time, you eventually would escape. In a thought experiment with nothing but Earth, the escape velocity for an unpowered body at any given distance from the center would be inversely proportional to the square root of that distance. At roughly 60,000,000 miles that velocity would be down to 200mph, so we could stop the engine at that point and escape ballistically. It would take roughly 30 years to get there, so it would not be very practical.

Jeff Root
2009-Jun-02, 07:04 AM
Suppose that I have a F/A-22 Raptor, airtight, with 2 machs at my
disposal, suppose also I have fairly good fuel reserves, can't I go
anywhere I want?
Taking my sweet time?

I wasn't talking about a conventional aircraft, this hypothetical craft
would be a spacecraft with the same powers, it wouldn't have turbofan
engines since there's no air up there.
What I am trying to visualize is; Is it possible to escape Earth's grip
with minimum speed, if the craft continues to climb, like, in a way
Space Shuttle, which has been it's highest orbit just to fix Hubble.
Your craft makes no sense. It isn't an airbreathing jet, and it isn't
a rocket. What is it? Exactly how much fuel does it carry?

A jet plane is able to fly a long distance and for a long time because
it has wings which provide lift. That lift keeps the plane up in the air.
Without the lift provided by the wings, the jet engines would need to
put out an enormous amount of thrust continuously to keep the plane
from falling, with some or all of the thrust directed downward instead
of just backward. That would use up fuel at a huge rate, so the plane
would not be able to stay up very long or fly very far.

A rocket works like a cross between a plane and a gun. Like a plane,
it carries the fuel it uses. Like a gun, it gets as much speed as it
needs, then stops burning the fuel and coasts the rest of the way.
Maximum efficiency is when all the fuel burns instantaneously, and
the rocket reaches the required speed immediatly. That has two big
problems: 1) The g-forces would destroy the rocket and its payload,
and 2) The atmosphere puts a lot of drag on anything moving through
it at high speed. So real rockets accelerate as quickly as possible,
and also try to get above the atmosphere as quickly as possible to
get past the drag problem, where they can then build up the speed
they need to get into orbit or escape the planet.

Without that speed, the rocket will not escape or go into orbit.
It will just fall back to Earth on an elliptical trajectory.

Your F/A-22 Raptor, with engines capable of operating in vacuum,
would be limited in altitude by the lack of air to provide lift. The
fact that the engines operate in vacuum means that it is a rocket,
carrying its own oxydizer in addition to fuel. Once the plane is at
an altitude where airbreathing engines no longer function and wings
no longer provide enough lift to hold the plane up, the rocket engines
need to start using that fuel and oxydizer at a high rate. So if you
want to get very far from Earth, you will need a huge amount of fuel
and a huge amount of oxydizer. Those require huge tanks. A very
big, heavy vehicle. Like the Space Shuttle.

If you could get your modified F/A-22 Raptor far enough from Earth
that Earth is no longer the dominant gravity source, you can escape
using either a gravitational slingshot or a more subtle maneuver, the
name for which I seem to have forgotten. The maneuver was only
developed within the last 25 years or so. It has been used by some
spacecraft to get to the Moon. I think the Japanese spacecraft now
orbiting the Moon (it will crash into the Moon in just a few days) may
have used the technique. Previous Japanese and American probes
have used it. It depends on a combination of the gravity of the Sun
and Moon to gradually pull a spacecraft away from Earth over many
orbits. So the spacecraft needs to be able to get into orbit, first.
A high orbit, at that. It is only useful in a limited range of situations.
Getting from high Earth orbit to Lunar orbit is one.

-- Jeff, in Minneapolis

stitt29
2009-Jun-02, 01:26 PM
Hi

I just want to jump in again with a small question. If we fire the projectile at between 8km/sec and 11 km/sec, it will move in an elliptical orbit. Is the centre of the earth(or close to it) one focus of that ellipse?

Hornblower
2009-Jun-02, 01:52 PM
Hi

I just want to jump in again with a small question. If we fire the projectile at between 8km/sec and 11 km/sec, it will move in an elliptical orbit. Is the centre of the earth(or close to it) one focus of that ellipse?

Yes, and it also is true for the parabolic or hyperbolic orbits that result from higher velocities. It is one of the "laws" of orbital motion, discovered by Kepler and explained gravitationally by Newton.

Jeff Root
2009-Jun-02, 07:44 PM
It has been said so many times, by so many people in so many different
venues, that the trajectory of a thrown object is a parabola, that it took
me a long time to be confident that they were all wrong -- the trajectory
is actually a portion of an ellipse. It is just that there isn't a whole lot
of difference between the portion of a parabola and the portion of an
ellipse that is above the ground. But it is neat to think that when I jump
six inches above the floor and am briefly in free-fall, I am actually orbiting
the Earth -- with a perigee close to the Earth's center, and apogee six
inches above the floor.

-- Jeff, in Minneapolis

solomarineris
2009-Jun-03, 03:09 AM
At roughly 60,000,000 miles that velocity would be down to 200mph, so we could stop the engine at that point and escape ballistically. It would take roughly 30 years to get there, so it would not be very practical.
Assuming you are correct with your calculations (which I believe you are).
You are saying it is possible.
Anyway, we are still stuck here, doesn't matter how deep we speculate.
Thanks for the education.