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

Janus
Staff Emeritus
Gold Member
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.

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.

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
Mentor
2020 Award
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.

mfb and mrspeedybob
jbriggs444
Homework Helper
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.

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!

Staff Emeritus
Just dig a tunnel straight down through the earth

Or use a big spring.

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
Mentor
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.

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Janus
Staff Emeritus
Gold Member
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.

Dale
sophiecentaur
Gold Member
2020 Award
Look up Newton's cannon.
But the cannon ball does not travel on a track and has to be travelling 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
Staff Emeritus
Gold Member
But the cannon ball does not travel on a track and has to be travelling 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
Gold Member
2020 Award
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.

*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.

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.

*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.

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.

Dale
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.

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
Homework Helper
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.

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.

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