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.