Trip to Space -- Can ship with 1g acceleration escape Earth?

[ckirmser]

In another forum, the question was raised, "could a ship with 1G acceleration escape the gravity well of a planet with 1G gravity?"

A popular response is, if the craft is aerodynamic, it could accelerate laterally until it reached escape velocity and then manage to get to space.

I don't believe this is a viable response, even if friction is ignored, but I really can't think of a clear, concise way to communicate this gut-feeling.

Am I right or is this actually a worthwhile answer?

Thanx!

 

[A.T.]

"could a ship with 1G acceleration escape the gravity well of a planet with 1G gravity?"

So the proper acceleration of the ship (measured with an on-board accelerometer) cannot be greater than 1g in magnitude?

Do you start from the surface, or at some distance where local gravity is 1g (the surface gravity being higher than 1g)?

If from surface, can you use it as support, while accelerating along it? Is the planet rotating?

 

[anorlunda]

If the ship can maintain constant acceleration no matter what the gravity, then every nonzero acceleration escapes.

Do you mean a ship with an engine or a thrown object with no engine and no acceleration after it is released?
When we talk about escape velocity it is the object case we think of.

 

[Janus]

In another forum, the question was raised, "could a ship with 1G acceleration escape the gravity well of a planet with 1G gravity?"

A popular response is, if the craft is aerodynamic, it could accelerate laterally until it reached escape velocity and then manage to get to space.

I don't believe this is a viable response, even if friction is ignored, but I really can't think of a clear, concise way to communicate this gut-feeling.

Am I right or is this actually a worthwhile answer?

Thanx!

Let's assume no atmosphere, a perfectly smooth and round planet, and your craft is resting on a cradle which rolls on a world circling track( to initially support the weight of your craft.)
You fire up the engine and your craft accelerates along the track. It will continue to gain speed until eventually it reaches orbital velocity, at which point the cradle will be no longer needed. As it continues to fire its engine, it will climb to a higher and higher orbit (it will spiral away from the planet). As it does so, it will trade speed for altitude. But this is okay, because as it gains altitude, the escape velocity that it needs to achieve also decreases. Eventually it will attain an altitude where its speed exceeds the escape velocity for that altitude.
It doesn't even matter how small an acceleration your craft has. Even if it could only generate 1/100 of a g, it still would eventually attain surface orbital speed, and then after that, slowly spiral away from the planet. The process would just be slower.

 

[russ_watters]

In another forum, the question was raised, "could a ship with 1G acceleration escape the gravity well of a planet with 1G gravity?"

A popular response is, if the craft is aerodynamic, it could accelerate laterally until it reached escape velocity and then manage to get to space.

I don't believe this is a viable response, even if friction is ignored, but I really can't think of a clear, concise way to communicate this gut-feeling.

Am I right or is this actually a worthwhile answer?

Thanx!

As is common with these Internet questions, wording and precision of assumptions matters a lot, but I think even heavily constrained, the answer is still yes:

If this is a craft with a thrust to weight ratio of 1.0 when standing still at sea level, as soon as it takes off and starts climbing, the thrust increases (rockets do better with less/no atmospheric backpressure) and the weight decreases (Newton's law of gravity), and vertical escape becomes possible.

 

[CWatters]

In another forum, the question was raised, "could a ship with 1G acceleration escape the gravity well of a planet with 1G gravity?"

Yes.

If it goes up vertically accelerating at 1g (=9.8m/s/s) then passengers inside will initially experience 2g. This will reduce asymptotically towards 1g as they get far from the planet. It will remain at 1g until the craft stops accelerating.

According to the table on Wikipedia...

https://en.wikipedia.org/wiki/G-force#Typical_examples

Standing on the Earth at sea level–standard = 1 g
Saturn V moon rocket just after launch = 1.14 g

If you can accelerate vertically from a planet then it doesn't actually matter what "g" is on that planet. For example if g was 23.572 m/s/s and the craft was able to accelerate vertically at 0.5g then it would be able to accelerate vertically at 0.5 * 23.572 = 11.786 m/s/s. Any number greater than zero will allow you to escape.

 

[ckirmser]

Well, I guess he's right then, though it seems intuitively that the vessel could never leave the gravity well. I'm thinking vectors and the total vector going radial to the center of the planet is still 1G and the ship would never be able to get a >1G opposing vector, since the craft is limited to 1G.

But, I must not be understanding the problem correctly - not outside the realm of possibility.

Thanx, guys!

 

[ckirmser]

So the proper acceleration of the ship (measured with an on-board accelerometer) cannot be greater than 1g in magnitude?

Yes. The ship can apply a 1G vector in any direction, but no more.

Do you start from the surface, or at some distance where local gravity is 1g (the surface gravity being higher than 1g)?

From the surface.

If from surface, can you use it as support, while accelerating along it? Is the planet rotating?

Well, sorta. The idea is that the ship uses aerodynamics to stay in the air until it gets up the speed to escape - according to the other guy, about Mach 33. And, yes, the planet is rotating.

 

[CWatters]

It's a bit of a trick question. It doesn't state under what conditions it can accelerate at 1g.

If it can accelerate vertically at 1g from the surface of a planet then that's one thing. If they meant it can only accelerate at 1g in free space then that's another. In the latter case it wouldn't even be able to lift off a 1g planet.

Edit: Actually that's not true. It could accelerate horizontally.

 

[ckirmser]

If the ship can maintain constant acceleration no matter what the gravity, then every nonzero acceleration escapes.

Really? I'm thinking vector addition here, so if one has a ship with an acceleration vector of 0° and a magnitude of 0.5G going against a gravity vector of 180° at 1G, how does the ship get past that 1G? I mean, when you add the vectors, you always get a negative 0.5G keeping the ship firmly planted on the surface.

Do you mean a ship with an engine or a thrown object with no engine and no acceleration after it is released?
When we talk about escape velocity it is the object case we think of.

This is a ship with an engine providing a constant 1G. So, I figure, the ship provides an acceleration away from the planet at 1G and the planet's gravity provides an opposite acceleration of 1G. This strikes me as providing a speed of 0 (1G-1G).

 

[CWatters]

This is a ship with an engine providing a constant 1G.

What do you mean by an engine providing a constant 1g? Think about the units.

 

[ckirmser]

It's a bit of a trick question. It doesn't state under what conditions it can accelerate at 1g.

If it can accelerate vertically at 1g from the surface of a planet then that's one thing. If they meant it can only accelerate at 1g in free space then that's another. In the latter case it wouldn't even be able to lift off a 1g planet.

It would be an acceleration of 1G in space, in an essentially gravity-free environment - given that nowhere is it totally "gravity-free." So, let's put it in this manner;

The ship has an acceleration of 9.8m/s/s.

You give the answer that I think it should be, that the gravity of the planet of 1G would negate the 1G acceleration of the ship. Even if it were to be on some track that circled the equator enabling it to build up speed. It just seems to me that the ship still has to get an acceleration of 1G+ radially away from the planet and that is simply not happening. But, if the ship reaches escape velocity on that track, what then? It's free from the need to accelerate, since it's already at escape velocity. All it would have to do is release its grip on the track and it would fling off to space.

But, that flies in the face of my gut saying, "that 1G acceleration is just not going to beat that 1G gravity, dadgummit!"

 

[ckirmser]

What do you mean by an engine providing a constant 1g? Think about the units.

9.8m/s/s.

 

[CWatters]

It would be an acceleration of 1G in space, in an essentially gravity-free environment

In that case I agree with you. It can't lift off vertically from a 1g planet.

 

[CWatters]

Even if it were to be on some track that circled the equator enabling it to build up speed. It just seems to me that the ship still has to get an acceleration of 1G+ radially away from the planet and that is simply not happening.

It would eventually be going very very fast (lot of KE) ... and then let go of the track.

 

[ckirmser]

It would eventually be going very very fast (lot of KE) ... and then let go of the track.

But, after letting go, would it fling off into space, at escape velocity+? Would this mean that a ship with <1G acceleration could also escape, it would just need to be on the track longer?

Doesn't this conflict with your prior comment where you agreed with me? Or, is "vertically" the qualifier that makes the difference? I mean, flying off the track when the ship reaches speed seems to make sense, but I can't get the vector addition issue out of my head. The acceleration is still only 9.8m/s/s. That's the most it can put into a direction radially away from the planet. But, the planet, exerting a 1G force in the opposite direction, would negate it.

Wouldn't it?

 

[jbriggs444]

But, after letting go, would it fling off into space, at escape velocity+? Would this mean that a ship with <1G acceleration could also escape, it would just need to be on the track longer?

Yes. As @Janus pointed out in #4 above.

However, if we require that the passengers never experience more than 1 g proper acceleration then this becomes problematic. The craft can never move from its starting point without either moving slightly downward or subjecting the passengers to a momentary acceleration slightly greater than 1 g.

[Which, on re-reading #16, may be your point]

 

[A.T.]

...flying off the track when the ship reaches speed seems to make sense, but I can't get the vector addition issue out of my head. The acceleration is still only 9.8m/s/s. That's the most it can put into a direction radially away from the planet. But, the planet, exerting a 1G force in the opposite direction, would negate it.

And what kind of trajectory does a mass follow when the forces on it cancel?

 

[ckirmser]

OK, I made a graphic to illustrate to myself what was going on and I think I see my confusion - sorta. In the images, the brown circle is the hypothetical "track" the ship rides on. Light Blue arrows indicate the vector for gravity, Red arrows represent the prior path carried forward to the next time block, Green arrows represent the acceleration vector and Black arrows represent the path traveled by the ship for the current time block. It's been too long since I had Calculus, so the direction of the acceleration vectors is just the starting position tangential to the track; I wasn't about to attempt to calculate the direction over the course of the track segment during the duration of the time represented.

Anywhoo, my intial take was to consider the Earth as a point source of gravity, so the acceleration of the ship to leave would be away from the point source and then, the gravity of Earth would cancel out the acceleration, giving a net velocity of 0;

But, the Earth, closer up, is a sphere. And, since the ship is on a track, the upward force of the track against the ship cancels out much of the force of gravity, until the point is reached where the gravity vector's magnitude fails to bring the actual path back down into contact with the track. But, by that time, the ship's velocity is enough to allow the further application of 1G acceleration to continue off-planet;

That is, if I've done this right - I think I have. But, still, my gut says it's wrong. The 1G of Earth negates the 1G of the ship, keeping the ship planet-bound. But, using the track seems to allow the ship to leave.

Although, I note that the Black actual paths, the first couple, anyway, intersect the Brown track. Should I have considered that this represents the ship coming to a stop at the point of the first intersection? If so, then my gut ends up being right and the 1G ship ain't a-goin' nowheres...

 

[Janus]

OK, I made a graphic to illustrate to myself what was going on and I think I see my confusion - sorta. In the images, the brown circle is the hypothetical "track" the ship rides on. Light Blue arrows indicate the vector for gravity, Red arrows represent the prior path carried forward to the next time block, Green arrows represent the acceleration vector and Black arrows represent the path traveled by the ship for the current time block. It's been too long since I had Calculus, so the direction of the acceleration vectors is just the starting position tangential to the track; I wasn't about to attempt to calculate the direction over the course of the track segment during the duration of the time represented.

Anywhoo, my intial take was to consider the Earth as a point source of gravity, so the acceleration of the ship to leave would be away from the point source and then, the gravity of Earth would cancel out the acceleration, giving a net velocity of 0;

View attachment 114033

But, the Earth, closer up, is a sphere. And, since the ship is on a track, the upward force of the track against the ship cancels out much of the force of gravity, until the point is reached where the gravity vector's magnitude fails to bring the actual path back down into contact with the track. But, by that time, the ship's velocity is enough to allow the further application of 1G acceleration to continue off-planet;

View attachment 114034

That is, if I've done this right - I think I have. But, still, my gut says it's wrong. The 1G of Earth negates the 1G of the ship, keeping the ship planet-bound. But, using the track seems to allow the ship to leave.

Although, I note that the Black actual paths, the first couple, anyway, intersect the Brown track. Should I have considered that this represents the ship coming to a stop at the point of the first intersection? If so, then my gut ends up being right and the 1G ship ain't a-goin' nowheres...

Look up Newton's cannon.

 

[Svein]

A thought: Once you escape from the Earth's gravity, if you accelerate at 1g for a year you will be at above 90% of the speed of light. Now you just have to invent such an engine...

 

[A.T.]

The 1G of Earth negates the 1G of the ship, keeping the ship planet-bound.

You should really look up Newton's Laws. When all the forces cancel, the ship will move along a straight line, so it cannot stay on a round planet.

 

[Janus]

A thought: Once you escape from the Earth's gravity, if you accelerate at 1g for a year you will be at above 90% of the speed of light. Now you just have to invent such an engine...

No, In one year ship time, you would only be up to 77% of c. In one year Earth time, you would be up to 71.7% of c.

 

[Svein]

No, In one year ship time, you would only be up to 77% of c. In one year Earth time, you would be up to 71.7% of c.

It may well be so. I just calculated it using Newtonian mechanics, assuming that the relativistic phenomena would not matter up to 90% of c.

 

[Janus]

It may well be so. I just calculated it using Newtonian mechanics, assuming that the relativistic phenomena would not matter up to 90% of c.

By 0.5c the time dilation factor is 0.866
at 0.717c it is 0.697
at 0.77c it is 0.638
at 0.9c it is 0.436

 

[Janus]

Although, I note that the Black actual paths, the first couple, anyway, intersect the Brown track. Should I have considered that this represents the ship coming to a stop at the point of the first intersection? If so, then my gut ends up being right and the 1G ship ain't a-goin' nowheres...[/QUOTE]
The black line are just an approximation of the path the ship would take if the track wasn't there (the actual paths would be curves). The track just prevents the ship from following that vector component of that path which points towards the center of the Earth. It provide the upwards force that cancels the Earth's gravity and allows the ship's rockets thrust to be applied to just increasing the speed of the ship.
You might try looking at it this way:
As the ship increases speed as it travels along the track, it is traveling in a circle. Objects traveling in a circle need a centripetal force acting on them in maintain the circular motion. (imagine the ship trying to travel in a circle in space far away from the Earth. It would need some inward acting force to keep it from flying off in a straight line.) For the ship on the track, gravity provides that centripetal force. At first, it is more than up to the task . But as the ships speed increases, the centripetal force needed to hold it to its circular path also increases. Eventually it will reach a speed where the Earth's gravity equals the magnitude of centripetal force needed. The ship has reached circular orbit speed.
The Earth's gravity is now just holding the ship to the circular path. Any further increase in the ship's speed means that the Earth's gravity is no longer enough to hold the ship to that circular path, and the ship will lift off the track and climb away from the Earth. As this happens, the Earth's gravitational pull also weakens. (if it wasn't for that fact, the whole concept of escape velocity wouldn't exist) This combination leads to the ship eventually leaving the vicinity of the Earth entirely.

 

[John Park]

But, that flies in the face of my gut saying, "that 1G acceleration is just not going to beat that 1G gravity, dadgummit!"

Reference https://www.physicsforums.com/threa...hip-with-1g-acceleration-escape-earth.906145/

Have you considered the so-called "centrifugal force"? Technically that's a fictitious force, but we can approach the same question in terms of the real centripetal acceleration, which is the inward acceleration that must be applied to keep a body in a circular orbit. For an orbit of radius r and a tangential speed v, it's a matter of geometry that the centripetal acceleration must be v²/r. For our purposes r is the radius of the Earth.

Your ship starts off motionless on the ground. The Earth provides a downward acceleration of g, but this is balanced by the normal reaction of the ground.

Now the ship accelerates horizontally and we ignore air resistance. As the ship builds up speed the required centripetal acceleration v²/r starts from zero and increases until it becomes equal to g. At this speed gravity is just providing enough acceleration to keep the ship in a circular orbit; the ship's weight, in terms of pressing on the ground, is zero. As the speed increases beyond this point the ship will lift off the ground. Initially it will be in an approximately circular orbit, but it will begin to spiral outwards and eventually escape. "Eventually" won't be very long in fact. A one-gee acceleration will get to escape speed in about twenty minutes.

 

[bahamagreen]

The ship is sitting on a pad at the Earth's surface.
The engine can provide 1G acceleration (which rating includes a full fueled ship) :)
Before starting, the ship rests on the pad, no motion.
When engaging the engine the ship's acceleration replaces the support of the pad.
Initially the ship's acceleration opposes and matches that of the Earth, so initially it floats steady...
The engine is rated at +G for the initial mass of the full fueled ship, but as fuel is consumed the ship gets lighter. :)
The ship starts to lift, and enjoys a decreasing Earth's G gradient all the way up.

 

[Dale]

However, if we require that the passengers never experience more than 1 g proper acceleration then this becomes problematic. The craft can never move from its starting point without either moving slightly downward or subjecting the passengers to a momentary acceleration slightly greater than 1 g.

Even in this case you could do it. Just dig a tunnel straight down through the earth, and accelerate downward with 1 g proper acceleration. Then after you pass the center of the Earth continue upward with 1 g proper acceleration. Right as you exit the surface you will have no coordinate acceleration, but you will go quickly past that spot and resume coordinate acceleration as you ascend.

 

[jbriggs444]

Even in this case you could do it. Just dig a tunnel straight down through the earth, and accelerate downward with 1 g proper

Agreed. The key being the "downward" bit.

 

[A.T.]

In essence:

If one interprets the conditions such that it can never start moving, then it trivially cannot escape.

But if it can start moving somehow, then any thrust for unlimited time can do unlimited work, so it can trivially escape.

 

[Thomas Wown]

Earths gravity is around 9 and most planes that take off Horizontally accelerate and 3-4 gz (Boeing). So yes it's possible. Common sense wins!

 

[Vanadium 50]

Just dig a tunnel straight down through the earth

Or use a big spring.

 

[ckirmser]

Well, it appears that, no matter how loud my gut screams at me, it's just flat wrong.

But, is it only because we're dealing with a sphere and the starting position of the ship is separated from the point source of gravity, the center of the Earth?
I mean, theoretically, if we deal with point sources of gravity and acceleration, then the "ship" has no track to ride and is forced, no matter what direction it points, to give its 1G acceleration directly against the 1G of gravity. The one cancels out the other and the ship goes nowhere.

I figure it must be because, as I said in an earlier post, the track is providing a virtual force opposite that of gravity, allowing the ship to get sort of a false acceleration upwards, even though all of its own acceleration is being used to push it along the track.

Still doesn't seem right...

 

[A.T.]

...if we deal with point sources of gravity...

What do you mean by that?

If you start at the point mass, then you have infinite Newtonian gravity, not 1g.

If you start at distance r from the point mass where gravity is 1g, then it's even simpler to escape with 1g proper acceleration, because the planet is not in the way.

 

[Nugatory]

I mean, theoretically, if we deal with point sources of gravity and acceleration...

Point sources of gravity are tricky because Newton's law of gravity applies in all the space around the point but not at the point itself. One way to see this is that at $r=0$ Newton's law of gravity gives you an infinite force pointing in no direction; the nonsensical result is the math's way of telling you that it doesn't work under those conditions. This is not a problem because such ideal point masses do not exist and it should be no surprise that the math does not describe the behavior of something that cannot exist. However, it does mean that if you try to analyze the behavior of the spaceship at $r=0$ you are committing a logical error similar to the one you are committing if you start a mathematical proof with something like "Let $n>1$ be a common factor of two different prime numbers $P$ and $Q$..."; the initial premise is bogus so can lead to bogus results.

the track is providing a virtual force opposite that of gravity, allowing the ship to get sort of a false acceleration upwards,

There is nothing virtual about that force. When the ship is resting on the track a spring scale between the ship and the track will clearly show that the track is pushing on the ship with some force; we infer that the force of gravity on the ship is equal to and opposite to the force of the track on the ship because the ship is not accelerating so the net force on the ship is zero. If the track were not exerting this upwards force on the ship it would fall towards the center of the earth. Nor is there any "false acceleration" involved; the $F$ in $F=ma$ is always the total force from all sources acting on the object that is accelerating with acceleration $a$. It pretty much has to be that way because the object in question only has one acceleration vector at any moment, no matter how many different forces are acting on it at that moment.

 

[Janus]

Well, it appears that, no matter how loud my gut screams at me, it's just flat wrong.

But, is it only because we're dealing with a sphere and the starting position of the ship is separated from the point source of gravity, the center of the Earth?
I mean, theoretically, if we deal with point sources of gravity and acceleration, then the "ship" has no track to ride and is forced, no matter what direction it points, to give its 1G acceleration directly against the 1G of gravity. The one cancels out the other and the ship goes nowhere.

I figure it must be because, as I said in an earlier post, the track is providing a virtual force opposite that of gravity, allowing the ship to get sort of a false acceleration upwards, even though all of its own acceleration is being used to push it along the track.

Still doesn't seem right...

Let's assume the mass of the Earth is concentrated to a point, but your ship still starts 1 Earth radius away where gravity is 1g. There is no track, but your ship is pointed in such a way that it is laying on its side. Without firing his engines, he would fall straight in towards the center. If he fires his engines at all, even briefly, he will put him self in an elliptical orbit around the central mass with the mass at one focus. The longer he fires his engine, the further his closest approach will be from the central mass.
Now even if he no longer fires his engines, he will pick up speed as he moves in closer to the center.
I mention this so we can consider the following scenario:
He fires his engines once briefly to put himself in that elliptical orbit. Once his reaches his closest approach he will picked up a great deal of speed. He will in fact be moving much faster than circular orbital speed for that same distance. Escape velocity is ~1.414 times the circular orbital velocity for the distance from the center we are considering. This means, that at this moment, our ship is closer to escape velocity than something in a circular orbit at that distance would be.
Now we make sure our ship is pointed in the same direction as our orbital motion, and fire our engines briefly again, increasing our speed even more. This changes the orbit, raising the other end of the orbit.
So let's say that this still left you short of escape velocity. Just wait until you've completed one whole orbit, and then do another brief thrust. This increases your velocity even more, pushing the high end even higher. Just keep repeating this until your last thrust puts you over escape velocity. This process works without your even having to provide thrust the entire time, and no matter how weak your thrust is. (This method is actually more efficient than continuous thrust.)

Orbital mechanics is one of those disciplines where intuitive "gut feeling" can often lead you badly astray.

 

[sophiecentaur]

Look up Newton's cannon.

But the cannon ball does not travel on a track and has to be traveling at the appropriate speed as soon as it is launched. It is a very different thing from a vehicle on a rail that could accelerate at any rate at all until the track ends, in principle. If you exchange the track for the lift from a wing, a similar argument applies and the vehicle can be accelerating continually whilst it has atmosphere to travel through. Drag is a factor of course but efficiency of a rocket system is the reason why they're operated with high acceleration.

 

[Janus]

But the cannon ball does not travel on a track and has to be traveling at the appropriate speed as soon as it is launched. It is a very different thing from a vehicle on a rail that could accelerate at any rate at all until the track ends, in principle. If you exchange the track for the lift from a wing, a similar argument applies and the vehicle can be accelerating continually whilst it has atmosphere to travel through. Drag is a factor of course but efficiency of a rocket system is the reason why they're operated with high acceleration.

The barrel of the cannon acts like the track from the time the powder is ignited until it leaves the muzzle and while the projectile accelerates. The track just serves the same purpose for a lower acceleration and longer run. The point is that it doesn't matter how the projectile achieves orbital/escape velocity relative to the surface, just that it eventually does.

 

[sophiecentaur]

The track just serves the same purpose for a lower acceleration and longer run.

Precisely. The OP asks about low values of acceleration. The cannon provides very high acceleration to get the same result.
What really counts is the Kinetic Energy that the Ship can be given and that can (at least in principle) be dished out in as long a time as you may choose.

 

[Photonboy]

*The main CONFUSION is the air resistance.

If we assume NO air at all, and assume that the ship can apply 1G maximum thrust then it can either accelerate directly UPWARDS just enough to cancel out out gravity (thus weighing NOTHING relative to a normal scale, but also not going anywhere). The ship can move sideways relative to the surface as well, but without any air it would just slide around on the surface.

Now, once we have AIR resistance this all changes. As long as the aerodynamics are sound, then we can fly like normal and eventually spiral off. It's actually impossible to have indefinite thrust as you lose mass but assuming that you'll eventually escape. You lose speed due to air resistance, but the air resistance also thins as you get higher so (in theory) you can escape. The thrust required isn't slightly more than 1G as it would be in a vacuum, plus since we are no longer at the 1G equalized point (if we assumed surface was where the vectors cancel to zero), we are higher.

As said elsewhere you need to be VERY SPECIFIC with the wording because having air, or wings etc changes the question.

 

[ckirmser]

What do you mean by that?

Well, typically, one doesn't do all the Calculus necessary to figure the gravity affecting a body on the surface of a planet by figuring out what it is from all the mass below the surface in all directions. Instead, one just assumes the gravity of the body is focused at the center of mass.

If you start at the point mass, then you have infinite Newtonian gravity, not 1g.

Not a point mass in fact, but only for the purpose of making the discussion and the math easier.

If you start at distance r from the point mass where gravity is 1g, then it's even simpler to escape with 1g proper acceleration, because the planet is not in the way.

Actually, using the Earth as an example, it's only 1g at the surface. At the center, it is 0g. So, all ships leaving Earth will start at r, which is, what, around 4,000 miles? At that distance, the effect of gravity is 1g.

 

[John Park]

*The main CONFUSION is the air resistance.

If we assume NO air at all, and assume that the ship can apply 1G maximum thrust then it can either accelerate directly UPWARDS just enough to cancel out out gravity (thus weighing NOTHING relative to a normal scale, but also not going anywhere). The ship can move sideways relative to the surface as well, but without any air it would just slide around on the surface.

But even on an airless world as long as the ship can accelerate horizontally it will eventually reach escape speed.

 

[John Park]

I figure it must be because, as I said in an earlier post, the track is providing a virtual force opposite that of gravity, allowing the ship to get sort of a false acceleration upwards, even though all of its own acceleration is being used to push it along the track.

That's more or less how I'd look at it.
Note that when space vehicles take off, they start out vertically until they're out of the atmosphere; then they shift to a roughly horizontal trajectory, which is apparently a more efficient way to build up speed.

 

[ckirmser]

Point sources of gravity are tricky because Newton's law of gravity applies in all the space around the point but not at the point itself. One way to see this is that at $r=0$ Newton's law of gravity gives you an infinite force pointing in no direction; the nonsensical result is the math's way of telling you that it doesn't work under those conditions. This is not a problem because such ideal point masses do not exist and it should be no surprise that the math does not describe the behavior of something that cannot exist. However, it does mean that if you try to analyze the behavior of the spaceship at $r=0$ you are committing a logical error similar to the one you are committing if you start a mathematical proof with something like "Let $n>1$ be a common factor of two different prime numbers $P$ and $Q$..."; the initial premise is bogus so can lead to bogus results.

Well, OK, true. I'm simplifying - perhaps over-simplifying - because this has only to do with a game. My reference is an old GDW game called Triplanetary.

In the game, 1 hex is equal to the distance traveled at 1g acceleration in the course of a turn. Final movement is determined by adding together the vectors of hexes entered and the acceleration of the ship for that turn. Planets are represented by a dot and each adjacent hex has a gravity arrow pointing towards the planet, like so;

A ship with 1g acceleration, trying to leave the planet, would go to an adjacent hex, add it's acceleration vector to the gravity vector of the planet, giving it a net movement of 0;

This is what I refer to as a "point source."

There is nothing virtual about that force.

I may have been a bit sloppy in my terms. By virtual, I meant that it is not a force of gravity or acceleration.

So, the ship can never leave the planet. But, it appears this is wrong by the use of a track around the planet - well, a theoretical track. Probably, in reality, there'd be no such track allowing the ship to get the speed up while being held down and having part of the gravity mitigated thereby.

 

[ckirmser]

Orbital mechanics is one of those disciplines where intuitive "gut feeling" can often lead you badly astray.

Kinda like vertigo.

"I know the bubbles are going that way, but they are WRONG! UP is over the other way!"

 

[jbriggs444]

In the game, 1 hex is equal to the distance traveled at 1g acceleration in the course of a turn.

These forums are about real world physics, not counter-factual game physics. In real life, 1 g is an acceleration, not a velocity.

 

[ckirmser]

As said elsewhere you need to be VERY SPECIFIC with the wording because having air, or wings etc changes the question.

I haven't been including air, just a track. Atmospherics adds more trouble to the main question.

However, my second image in post #19 is composed just from the vectors given by gravity and acceleration, and the supportive force of the track. No air at all.

 

[ckirmser]

These forums are about real world physics, not counter-factual game physics. In real life, 1 g is an acceleration, not a velocity.

The game example was just to illustrate how I was coming at this. It is still vector math, be it real life or game theory. And, in the game, the vector is an acceleration, not a velocity.

When extrapolated to a sphere, and the same vector math applied, I was able to visualize far better why the ship could leave when running on a track.

 

[sophiecentaur]

That's more or less how I'd look at it.
Note that when space vehicles take off, they start out vertically until they're out of the atmosphere; then they shift to a roughly horizontal trajectory, which is apparently a more efficient way to build up speed.

Efficiency is what it's all about. The OP is about 'Acceleration' and I'm afraid that a discussion which doesn't include the Energy requirements is really hardly worth having when space launches are concerned.