If a rocket ascends at 1 g, does it feel 2 g?

[DrZygote214]

My intuition says yes because we are always feeling 1 g just sitting here, so a rocket ascending at say 1 g, has to feel 2 g right?

But when does this stop being the case? A rocket in orbit is in free fall (weightless), so it feels no g. So then if it turns on its engine and accelerates at 0.25 g, the rocket feels 0.25 g.

So at some point between launch and achieving orbit, somehow "cumulative gravity", for lack of a better term, disappears? My intuition says it disappears gradually as the flight angle changes from 90 degrees to 0 degrees. Too bad i cannot test this myself.

For the sake of concreteness, assume the rocket is moving straight up away from the ground. In terms of actual motion, it is moving at 1 g, so after 1 second it will be 4.905 m above ground, after 2 seconds it will be 19.62 m, after 3 s it will be 44.145 m. It feels 2 g?

Yes i tried to google this with various phrasing, but always got results about old 1g/2g/3g phones.

 

[exchemist]

I think you are right. But, surely, if the rocket thrust were to remain constant, it would produce a thrust of 2g at whatever angle the rocket is, wouldn't it? So the astronauts would feel a 2g combined weight/acceleration, in the direction of the nose of the rocket, all the way up, until the motor was cut. Or so it seems to me.

 

[Farsight]

...So at some point between launch and achieving orbit, somehow "cumulative gravity", for lack of a better term, disappears?

You can simplify this by thinking of a rocket which goes straight up, and doesn't attempt to go into an orbit. You're the astronaut. When you climb aboard the rocket on the launchpad, you're feeling 1g. When the rocket takes off with a 1g acceleration you're feeling 2g. However when you're a million miles from Earth, the Earth's gravity is hardly detectable. If you dropped a brick from there, it doesn't fall downwards at 9.8m/s/s. So you don't need a 2g acceleration to get away from the Earth any more. You can reduce the thrust of the rocket engines. Then when you're further away, you can reduce them further, and further, and further. In the end you can turn them off, because you've got escape velocity. The Earth will slow you down a little, but never enough to pull you back to Earth.

 

[origin]

So at some point between launch and achieving orbit, somehow "cumulative gravity", for lack of a better term, disappears? My intuition says it disappears gradually as the flight angle changes from 90 degrees to 0 degrees. Too bad i cannot test this myself.

Here is a site that will help you.

For the sake of concreteness, assume the rocket is moving straight up away from the ground. In terms of actual motion, it is moving at 1 g, so after 1 second it will be 4.905 m above ground, after 2 seconds it will be 19.62 m, after 3 s it will be 44.145 m. It feels 2 g?

The g force would be the acceleration of the rocket plus the accleration of gravity which will decrease with the following equation:

$$g_f = g_o (\frac{r_e}{r_e + d})^2$$
$$g_f$$ acceleration of gravity at distance d
$$g_o$$ acceleration of gravity at the surface of earth
$$r_e$$ radius of the earth
$$d$$ distance from the surface of the earth.

You can see at large distances from earth the $$r_e$$ term is negligable.

 

[Janus58]

You can simplify this by thinking of a rocket which goes straight up, and doesn't attempt to go into an orbit. You're the astronaut. When you climb aboard the rocket on the launchpad, you're feeling 1g. When the rocket takes off with a 1g acceleration you're feeling 2g. However when you're a million miles from Earth, the Earth's gravity is hardly detectable. If you dropped a brick from there, it doesn't fall downwards at 9.8m/s/s. So you don't need a 2g acceleration to get away from the Earth any more. You can reduce the thrust of the rocket engines. Then when you're further away, you can reduce them further, and further, and further. In the end you can turn them off, because you've got escape velocity. The Earth will slow you down a little, but never enough to pull you back to Earth.

That's a wasteful way of doing it. You want to burn all the fuel you need to achieve escape velocity while you are as close to the Earth as possible. If 2g is the maximum thrust you can maintain, you burn your engines at this thrust until you reach escape velocity. You will feel 2g of force of force the whole time, just like exchemist said. Less and less of your thrust will be used to fight gravity and more and more will go to accelerating the rocket. If the rockets cut out, you go instantly weightless, regardless of whether or not you have achieved escape or orbital velocity.

 

[Farsight]

That's a wasteful way of doing it..

No problem. I was just trying to simplify the situation with a straight-up rocket and no orbit.

 

[Aqueous Id]

You can simplify this by thinking of a rocket which goes straight up, and doesn't attempt to go into an orbit. You're the astronaut. When you climb aboard the rocket on the launchpad, you're feeling 1g. When the rocket takes off with a 1g acceleration you're feeling 2g. However when you're a million miles from Earth, the Earth's gravity is hardly detectable. If you dropped a brick from there, it doesn't fall downwards at 9.8m/s/s. So you don't need a 2g acceleration to get away from the Earth any more. You can reduce the thrust of the rocket engines. Then when you're further away, you can reduce them further, and further, and further. In the end you can turn them off, because you've got escape velocity. The Earth will slow you down a little, but never enough to pull you back to Earth.

You meant 1g acceleration, which is given in the question.

Will a rocket accelerating at 1g reach one million miles? What is the minimum acceleration needed to reach 1 million miles, given the ideal conditions of maintaining exactly constant acceleration, once the launch begins?

 

[OnlyMe]

That's a wasteful way of doing it. You want to burn all the fuel you need to achieve escape velocity while you are as close to the Earth as possible. If 2g is the maximum thrust you can maintain, you burn your engines at this thrust until you reach escape velocity. You will feel 2g of force of force the whole time, just like exchemist said. Less and less of your thrust will be used to fight gravity and more and more will go to accelerating the rocket. If the rockets cut out, you go instantly weightless, regardless of whether or not you have achieved escape or orbital velocity.

Wasn't the question phrased in terms of uniform acceleration, rather than uniform thrust?

If you maintain a uniform acceleration straight up from the ground, you would begin by feeling 2g acceleration, adding your velocity to the initial effect of the earth's gravity. But if you are maintaining a uniform acceleration, relative to your starting location on the surface of the earth, the earth's contribution to what you feel drops off with distance, as does the energy or force required to maintain an acceleration of 1g.

As you restated the question, burning your fuel at a constant and uniform rate, your conclusion is correct, but it changes the OPs question from one of acceleration, to one of energy expenditure, even what best method to escape earth's gravity might be. Valid but not quite what it seems the OP was questioning.

 

[danshawen]

If a downward thrust of 1g acts to propel a rocket upward, another 1g to make it move upward is not necessary unless you wish the rocket to ascend at a rate as fast as it would otherwise fall. Unless it is something like a moon shot, a deep space probe, or is launching a geostationary satellite payload, it need only accelerate fast enough and long enough for it to reach orbital velocity at some pre determined altitude.

If you are standing on the top of a very tall tower, even one at a low orbital distance from the surface of the Earth, you will still experience an appreciable amount of Earth gravity. Note, you are NOT orbiting: just standing on high altitude but geostationary terrain. The force of gravity varies as the inverse square of the distance, so standing on a tower 4000 miles high (r' = 2r), your weight would still be fully 1/4 what it was on the surface.

 

[origin]

The force of gravity varies as the inverse square of the distance, so standing on a tower 4000 miles high (r' = 2r), your weight would still be fully 1/4 what it was on the surface.

That is not quite correct since the earth is not a point source at that distance. You would have to use the equation that I provided in post 4.

 

[DrZygote214]

You guys make a good point that earth's g diminishes with distance , so it is not always 9.81 m/s/s. However, at orbital altitudes of around 300 km above MSL, g = G * 5.972 e24/(6371000+300000)^2 = 8.96 m/s/s.

So it's not negligible, but not a huge difference either.

OnlyMe also makes a good point about uniform acceleration. In truth, rockets do not maintain such a thing. Each stage tries to maintain uniform thrust, but this does not translate to uniform acceleration, because their mass decreases as fuel is used up. And if we're being really accurate, a rocket's stage does not really maintain uniform thrust either.

My OP was about uniform acceleration and about getting into orbit, not escape velocity. Maybe a scenario with escape velocity could help us understand something, idk, but I'm gonna stick to orbit velocity.

Let's just assume the launching craft has constant mass and constant thrust. mass = 1,000 kg and thrust = 9,810 Newtons. This will produce a = 9.81 m/s/s, and it will be constant acceleration (at first). That's numerically equivalent to 1 g but it's important to distinguish a from g. a is the acceleration of the actual motion. That's why in my OP i gave an example with time so we see the actual motion of the craft. g is the g-force felt by the rocket and any occupants.

Since we are all feeling 1 g right now just sitting still, then the rocket feels a force of 2 g at first, right? I mean, that has to be true...? but I can't seem to prove it in my mind!

It goes straight up, but later it tilts towards a path parallel to the ground so that it gets into orbit. Somehow during this transition from straight up flight to parallel flight, "cumulative g-force" disappears.

I think we need to do this looking at the y-component and x-component separately. Argh, I tried to do all that vector stuff in one sitting but couldn't. I'll make another post once I succeed.

 

[billvon]

Will a rocket accelerating at 1g reach one million miles? What is the minimum acceleration needed to reach 1 million miles, given the ideal conditions of maintaining exactly constant acceleration, once the launch begins?

Any acceleration at all will result in you reaching one million miles eventually. (Note that the acceleration you FEEL will be 1G plus any actual acceleration, so if your G meter says 1G at launch you're not going anywhere.)

 

[origin]

You guys make a good point that earth's g diminishes with distance , so it is not always 9.81 m/s/s. However, at orbital altitudes of around 300 km above MSL, g = G * 5.972 e24/(6371000+300000)^2 = 8.96 m/s/s.

So it's not negligible, but not a huge difference either.

OnlyMe also makes a good point about uniform acceleration. In truth, rockets do not maintain such a thing. Each stage tries to maintain uniform thrust, but this does not translate to uniform acceleration, because their mass decreases as fuel is used up. And if we're being really accurate, a rocket's stage does not really maintain uniform thrust either.

My OP was about uniform acceleration and about getting into orbit, not escape velocity. Maybe a scenario with escape velocity could help us understand something, idk, but I'm gonna stick to orbit velocity.

Let's just assume the launching craft has constant mass and constant thrust. mass = 1,000 kg and thrust = 9,810 Newtons. This will produce a = 9.81 m/s/s, and it will be constant acceleration (at first). That's numerically equivalent to 1 g but it's important to distinguish a from g. a is the acceleration of the actual motion. That's why in my OP i gave an example with time so we see the actual motion of the craft. g is the g-force felt by the rocket and any occupants.

Since we are all feeling 1 g right now just sitting still, then the rocket feels a force of 2 g at first, right? I mean, that has to be true...? but I can't seem to prove it in my mind!

It goes straight up, but later it tilts towards a path parallel to the ground so that it gets into orbit. Somehow during this transition from straight up flight to parallel flight, "cumulative g-force" disappears.

I think we need to do this looking at the y-component and x-component separately. Argh, I tried to do all that vector stuff in one sitting but couldn't. I'll make another post once I succeed.

Every thing you need is in the link I gave you. Go about 1/3 the way down the article to Launch of a Space Vehicle and read that.

 

[James R]

DrZygote:

The acceleration you feel inside the rocket is not necessarily the same as the acceleration of the rocket (relative to earth). What you feel depends on the force that the rocket exerts on you at any instant of time.

Sitting on the launch pad, the rocket pushes up on you such that you feel a force equal to your normal weight and equivalent to what you would feel if the rocket was accelerating upwards in free space (far from Earth) at 1 g.

When the rocket takes off an accelerates upwards at 1 g relative to the ground, then you feel twice your normal weight, equivalent to what you would feel if the rocket was accelerating in free space at 2 g.

When the rocket reaches orbit, then the Earth's gravity is still accelerating it at almost 1 g, all the time. But you are also being accelerated by the Earth at 1 g, all the time. So, what you feel is zero weight. The rocket and you are both accelerating towards the Earth at the same rate, so you feel weightless. That 1 g from the Earth is still there when you're sitting in the International Space Station (slightly reduced, but not by much), but you don't feel it.

 

[Aqueous Id]

Any acceleration at all will result in you reaching one million miles eventually. (Note that the acceleration you FEEL will be 1G plus any actual acceleration, so if your G meter says 1G at launch you're not going anywhere.)

Yes, that's correct. Any amount of constant acceleration (a>0) is sufficient to reach escape velocity. That amount a will feel like g+a on the launch pad at the instant of liftoff and it will decrease with altitude as Origin said. And as Janus58 pointed out, other practicalities require a fast initial burn, meaning constant acceleration is impractical. And as JamesR said, for both of the above reasons, the crew will feel a varying amount of force.

My goal is to check Farsight for saying "you won't need a 2g acceleration to get away from the Earth any more". The OP asked us to assume a=g, not a=2g. The point here, as you said, is that you only "need" a>0. My quibble with him is over that word "need" as well as the minimum amount of acceleration "needed".

Evidently Farsight is still struggling with reference frames. He wants to add g to a (given a=g) to claim the ship is actually accelerating at a = 2g, believing this is a condition needed to escape the Earth's gravity.

What Farsight should have said is that the condition of escaping Earth's gravity occurs at a particular velocity, not a particular acceleration.

That particular velocity -- as you and most other readers well know -- is called the escape velocity, namely, when the kinetic energy of the vehicle is greater than the work done by gravity. And of course when they are equal then orbit occurs. When the kinetic energy is less than the work done by gravity then the vehicle is descending. You know all of this already. And why is that? Because you actually studied physics! Proving once again that science and academia are correct and Farsight is wrong.

Also proving that people who took physics tend to agree on facts as a matter of being correct rather than consensus. Of course the trivial answer is that there is consensus in science that it's no good to be wrong.

Just to clear the air, since Farsight has decided that one of his goals in life is to sow confusion and conflict into this forum. And since he insists that he is always right and the experts are always wrong.

Thanks for your posts. They are a great antidote for anyone who is sick of Farsight's.

 

[DrZygote214]

I think I finally figured it out. It happened as I was drawing the diagrams with acceleration arrows. I drew a point mass of 1,000 kg, a down arrow labeled -9.81 and an up arrow labeled +9.81.

In that utterly ridiculous moment I realized that these two accelerations canceled each other out and the craft has no net acceleration. Therefore, I had to draw the up arrow twice as large: +19.62 in order to get a net acceleration of +9.81.

So it hit me. You do feel 2g's, but only in reaction to your thrust and not gravity. Man I cannot stress to myself how dangerous this was, getting the right answer but for the wrong reasons.

I remembered from high school something called the normal force. I hate this term because there's nothing normal about it, but basically, the g-force you are feeling right now is not gravity, but rather the ground pushing you up. If the ground did not produce this counterforce, then gravity would be the only force and you would be falling right now. But thankfully, the "normal" force is here to give a net accel of zero.

So nothing actually feels gravity! (unless its a tidal force, dont get me started). That's why a probe can slingshot around Jupiter and its accelerometers will read zero the whole time. Come to think of it, that's really the same thing as orbiting Earth. The ship is just slingshotting around again and again, weightless, so accelerometers read zero.

The way I reconcile this in my head is that gravity acts on every molecule of your body. So you do not feel squished or pulled as gravity acts. Every molecule gets pulled at the same rate and so they do not move relative to each other, therefore no squishing feeling.

 

[DrZygote214]

I forgot to add about the transition from straight up flight to parallel flight. There actually is no transition in g-forces. Assuming the mass is constant 1 ton and the thrust is constant 16,920 newtons, you will always feel 2 g's regardless of what direction you're pointing.

However, starting out straight up your actual motion will be accelerating up at 9.81 m/s/s (in the y axis). Later, once you turn and fly parallel to the ground, your actual motion will be 19.62 m/s/s (in the x axis).

During this time, you are "falling" parabolically in the y axis. An orbital rocket will ascend so fast that it accumulates enough upward motion so that it can coast in the y axis with enough time to spare so that it can propel you to orbital speed in the x axis. By the time it reaches a peak in the parabola, it has already gotten enough thrust in the x axis to get into orbit.

BTW thanks for posting that link to http://www.braeunig.us/space/orbmech.htm. I remember that website and used it a lot for the chemical part of rocket science. Never saw the section for orbital mechanics. It is one of the best sites for this sort of thing, very compact and accessible, not ad-infested, and doesnt waste your time with "professional fluff" such as forewords.

 

[Billy T]

Some observations on the OP's “two step” process to orbit from an energy POV (by physicist who likes to think, but not to calculate):

As the square of the period, P, is proportional to the cube of the orbit radius, R, for circular orbit, and P is 2(pi)R/s where s is the speed, we have:
(R/s)^2 ~ ~ R^3 or R(s^2) is a constant.

As the kinetic energy, Em is ~ ~ s^2 (Called Em with m for motion, in contrast to Eg gravitational energy), we have REm is a constant or Em = K/R where K is that constant.

Defining the zero of Eg as when R is very large, and noting, so as to not confuse, that Eg goes as inverse R, (not R^2) and the only difference in Eg are independent of the arbitrary choice of where Eg = 0, we have Eg ~ ~ {(1/r) - (1/R)} which is positive as “r” is a radius closer to the center of the Earth than R or Eg = k{(1/r) - (1/R)} which is the gravitational energy needed to go from r to R, where k is another positive constant. In the case of interest, r = Re, the radius of the earth's surface.

For circular orbits Em is always equals, 0.5 |Eg| or Em = K/R = 0.5k{(1/Re) - (1/R)}

The total energy required, for orbit at R (neglecting losses to air friction in the climb out of the atmosphere), Et = Em* + Eg = (0.5 + 1)k{(1/Re) - (1/R)} = 1.5k{1/Re – 1/R}.

This is obviously less when the 2nd negative term makes the greatest reduction – i. e. as one should have known, less energy is required to go into low earth orbit, even by this "two step" route, but this math quantifies it (if you are willing to evaluate k)

However, if the OP, wants the climb up to always be along same radial line from the center of the earth, more energy would be required for continuous lateral thrusting during the climb up to R. If it is not supplied, the circumferential speed of the launch pad will be conserved, and as altitude increases the angular rate of changes decreases. I. e. from POV of the launch pad observer, the rocket is curving to the west as it climbs, not going radial straight up. Or in other words, that conserved speed is not enough to go around the earth at any of the increasing Rs from the earth in 24 hours.**

* If you launch at local midnight from the equator, part of the Em required is “free” due to fact earth's surface spin speed there (~1000mph as I recall) is added to its orbital speed, which is max at perigee (a winter day for the Northern hemisphere as I recall).

** BTW this is also true for the rigid vertical "space elevator" structure. If no lateral thrusting is provided and it extends "straight up" (along one constant earth radius), there will be torques trying to bend it into this westward curve. Perhaps it should just be a long westward curving cable with high-altitude, low pressure H2 filled, "balloon kite" at the end? - I have not thought much about this, but that is my "first guess." Might be good way to collect data on primary cosmic rays. I am sure there are weeks of reading of the analysis papers that address the space elevator's bending torque problem. I just have no interest in reading them.

 

[OnlyMe]

I think I finally figured it out. It happened as I was drawing the diagrams with acceleration arrows. I drew a point mass of 1,000 kg, a down arrow labeled -9.81 and an up arrow labeled +9.81.

In that utterly ridiculous moment I realized that these two accelerations canceled each other out and the craft has no net acceleration. Therefore, I had to draw the up arrow twice as large: +19.62 in order to get a net acceleration of +9.81.

When you are standing on the ground at the launch pad, you are experiencing 1g of acceleration, which is equivalent to +9.81m/s/s acceleration, as per the equivalence principle and the local effect of gravity. If you accelerate up from that location at +9.81 m/s/s you then feel 2g or +19.62 m/s/s acceleration. The down label of -9.81 m/s/s is an error. Gravitation pulls you toward the center of mass, but it's equivalent in acceleration is away from the center of mass.