Escape velocity for rocket launches

[jablonsky27]

escape velocity is defined in resnick,halliday,walker as 'the certain minimum initial speed that will cause it to move upward forever, theoretically coming to rest only at infinity. This initial speed is called the escape speed.'

the way escape speed is defined here gives me the impression that it is applicable only to projectiles which are given an initial thrust. during their entire trip after launch, they are not propelled.

if it is really the case then why do we need to worry about escape velocity for rocket launches? what stops a rocket traveling at say 2000km/h vertically upwards from the ground from leaving the Earth's gravitational field? afterall during its entire flight, it is being propelled upwards constantly.

 

[dst]

escape velocity is defined in resnick,halliday,walker as 'the certain minimum initial speed that will cause it to move upward forever, theoretically coming to rest only at infinity. This initial speed is called the escape speed.'

the way escape speed is defined here gives me the impression that it is applicable only to projectiles which are given an initial thrust. during their entire trip after launch, they are not propelled.

if it is really the case then why do we need to worry about escape velocity for rocket launches? what stops a rocket traveling at say 2000km/h vertically upwards from the ground from leaving the Earth's gravitational field? afterall during its entire flight, it is being propelled upwards constantly.

Absolutely nothing stops it, provided it can keep a minimum speed that isn't zero.

The "escape velocity" applies only to cases of initial thrust, with no acceleration afterwards - to my knowledge.

 

[jablonsky27]

exactly my thoughts dst. but again, if that is really the case then why do we need to worry about the 11km/s figure during rocket launches? those multistage boosters are there just to achieve that speed.

 

[rcgldr]

It takes less fuel overall if the acceleration is done quicker. For human spacecraft , the limit is around 3 to 4g's, but for satellites, it can be much higher, although aerodynamic drag limits the maximum amount of efficient acceleration possible.

One way of optimzing fuel usage is to use the centrifugal inertial reaction of the rocket to oppose gravity, instead of using thrust alone to reach high orbital altitudes, the moon, or other planets.

 

[jablonsky27]

hi jeff. You are talking about acceleration, i want to know about escape velocity.. I understand that there is a relation between fuel consumption and acceleration but that just tells you how fast or slow you should reach the escape velocity.
My question is why the velocity? Why can't a rocket leave Earth at a constant speed(which is less than the escape speed) and continuous thrust and still be able to reach moon?

 

[rcgldr]

Why can't a rocket leave Earth at a constant speed(which is less than the escape speed) and continuous thrust and still be able to reach moon?

It could, but it would require a huge amount of fuel. In the worst case scenario, imagine a rocket just hovering, consuming fuel and not going anywhere.

 

[aerospaceut10]

The idea of the escape velocity is that when the thrust is applied, it's supposed to be seen as a...very short impulse. If you delivered all your fuel, theoretically, at one time, and managed to get an impulse to have your spacecraft reach that escape velocity, then that ensures the craft to escape.

However as you can see most spacecraft don't do that currently and it's a bit more drawn out with the thrust. But in most spacecraft maneuvers when they change orbits, planes of orbits and such, they do short impulses for delta V.

 

[jablonsky27]

wikipedia says,

Planetary or lunar escape velocity is sometimes misunderstood to be the speed a powered vehicle (such as a rocket) must reach to leave orbit; however, this is not the case, as the quoted number is typically the surface escape velocity, and vehicles never achieve that speed direct from the surface.

In fact a vehicle can leave the Earth's gravity at any speed. At higher altitude, the local escape velocity is lower. But at the instant the propulsion stops, the vehicle can only escape if its speed is greater than or equal to the local escape velocity at that position- at sufficiently high altitude this speed can approach 0 m/s.

so i guess the speed and acceleeartion that the rocket has is for optimisation of fuel consumption.

 

[russ_watters]

exactly my thoughts dst. but again, if that is really the case then why do we need to worry about the 11km/s figure during rocket launches? those multistage boosters are there just to achieve that speed.

what stops a rocket traveling at say 2000km/h vertically upwards from the ground from leaving the Earth's gravitational field?

There aren't many practical uses for a rocket that travels at 2,000 km/h. That isn't fast enough to orbit and it would take around 4 years just to reach Mars.

We don't ever need to deal with the 11km/s figure itself, but low Earth orbit requires around 8km/s.

 

[pallidin]

'There aren't many practical uses for a rocket that travels at 2,000 km/h. That isn't fast enough to orbit..."

With due respect, isn't that part of the statement in contradiction to the Wiki article which says
"Planetary or lunar escape velocity is sometimes misunderstood... In fact a vehicle can leave the Earth's gravity at any speed."

 

[jablonsky27]

..rocket that travels at 2,000 km/h. That isn't fast enough to orbit..

isnt the orbital speed inversely dependant upon the arbital altitude?

and is there any relation between orbital speed and escape velocity?

 

[pallidin]

Hello. Escape velocity is a locality term. It has no other meaning.

 

[jablonsky27]

what does that mean? 'locality term'?

 

[D H]

Hello. Escape velocity is a locality term. It has no other meaning.

While escape velocity is indeed a function of distance from the gravitational body, it has a lot of meaning. The escape velocity (a bit of a misnomer; it should be escape speed) at a given altitude is the speed of an object on a parabolic orbit at that altitude. If the object's speed exceeds the escape velocity it is on a hyperbolic trajectory. Smaller speeds mean the object is in an elliptical orbit.

Why can't a rocket leave Earth at a constant speed(which is less than the escape speed) and continuous thrust and still be able to reach moon?

Theoretically, yes. Practically, no. The amount of fuel required to do so would be enormous.

wikipedia says,

Planetary or lunar escape velocity is sometimes misunderstood to be the speed a powered vehicle (such as a rocket) must reach to leave orbit; however, this is not the case, as the quoted number is typically the surface escape velocity, and vehicles never achieve that speed direct from the surface.

In fact a vehicle can leave the Earth's gravity at any speed. At higher altitude, the local escape velocity is lower. But at the instant the propulsion stops, the vehicle can only escape if its speed is greater than or equal to the local escape velocity at that position- at sufficiently high altitude this speed can approach 0 m/s.

so i guess the speed and acceleeartion that the rocket has is for optimisation of fuel consumption.

The initial work done by mission planners assumes "impulsive burns" -- that is, instantaneous changes in velocity. Say we want to plan a mission to Jupiter with flybys of Venus, Earth, and Earth again for gravity assists, with a final burn to place the vehicle in orbit around Jupiter. (This is what NASA's Galileo mission did; it is called a Venus-Earth-Earth gravity assist trajectory, or VEEGA for short.) Impulsive burns are needed at Earth to escape Earth's gravitational field and place the vehicle on an Earth-Venus trajectory, sometime before each flyby to set the vehicle up for the next phase, and at Jupiter to place the vehicle in orbit. Summing up the magnitudes of the changes in velocity gives the total delta V for the mission. This total delta V is closely aligned with the fuel needed to achieve the mission.

There is a slight problem with working in terms of impulsive burns: Doing so requires infinite thrust (aka infinite impulse). Real vehicles cannot perform impulsive burns. After the initial work, mission planners have to do a reality check because the finite impulse burns used on real vehicles inherently requires more energy to achieve the same change in velocity than does using an impulsive burn.

We use finite impulse burns for two reasons.

• We can't achieve the infinite thrust needed for impulsive burns.
• If we could, doing so would kill the passengers or crush the satellite.

The acceleration of a rocket is the rocket's thrust divided by it's mass. The initial acceleration is fairly low because the rocket is loaded with fuel. The acceleration increases as the rocket consumes fuel. This increasing acceleration can become too much for the passengers or cargo to withstand; see the second item on why we don't use impulsive burns. The Shuttle has to throttle down the engines as it burns fuel to keep thrust below three g.

 

[russ_watters]

'There aren't many practical uses for a rocket that travels at 2,000 km/h. That isn't fast enough to orbit..."

With due respect, isn't that part of the statement in contradiction to the Wiki article which says
"Planetary or lunar escape velocity is sometimes misunderstood... In fact a vehicle can leave the Earth's gravity at any speed."

isn't the orbital speed inversely dependant upon the arbital altitude?

and is there any relation between orbital speed and escape velocity?

The critical word in that first sentence was the word practical (which should be obvious from the sentence that follows it).

I don't feel like calculating the actual altitude right now (at the very least, it is well beyond the orbit of the moon), but what use would a satellite orbiting at such a high altitude be?

In fact, you really can't get to the moon (to land safely on it) at 2,000 km/hr, since the moon orbits at 3,600 km/hr. If you flew straight up at 2,000 km/hr, you'd still have to accelerate in order to land on the moon.

So as a practical matter, most of our rockets end up in low Earth orbit, at 250-500 km up. Most of the energy involved in an orbit at that altitude is from the speed, so rockets pitch over very soon after they lift off, spending most of their time/fuel trying to reach the necessary speed (~7km/sec).

For a rocket really trying to leave Earth and head for deep space, where is it going? Sure, you could leave Earth at a constant 2,000 km/h, but as a practical matter, trips to the inner solar system would take too long and trips to the outer solar system would take generations.

I'm sure NASA scientists/engineers work hard to find the optimal acceleration rate, but the reality is that escape velocity is an energy thing: whether you fly out at a small but constant speed or get shot out of a cannon at escape velocity, the amount of potential energy to be overcome is the same. The difference with flying out at a slow speed is that you also need a constant 1g (whatever 1g is at your altitude) of acceleration force to keep your constant speed.

 

[rewebster]

jablonsky27-- just checking if you are confusing 'escape velocity' with 'orbital velocity'?

 

[HallsofIvy]

jablonsky27-- just checking if you are confusing 'escape velocity' with 'orbital velocity'?

No, he's not. The definition in his original post was clearly for "escape velocity".

 

[HallsofIvy]

There was a rather silly television program a number of year ago (I forget the name but it starred Andy Griffith, for God's sake!) about a junk dealer who was going to make a fortune by going to the moon to recover material left there by the Apollo missions. One key point was that he was going to do this "cheaply" by accelerating very slowly: they did one demonstration in a car in which they accelerated slowly but long enough to reach speeds that frightened the person being demonstrated to! The program deteriorated from there with mobsters and spies every week.

In addition to other problems with "start slowly" already mentioned, one important reason to reach as fast a speed as you can as quickly as you can is to get through the atmosphere as quickly as possible.

 

[bkelly]

This is for jablonsky27

I don’t see a date on any of these posts, so this might be too late. But what the heck.

Here is a layman’s answer using the Earth as an example.

At any given altitude from the Earth there is an escape velocity. If an object is at any altitude, and its velocity is equal to or greater than the escape velocity at that altitude, then it will escape the gravitation pull of the earth. No further propulsion needed.

The further from Earth you start, the lower the escape velocity.

If you want to escape the Earth with a lower speed you must provide continuous thrust until your velocity equals the escape velocity. That means you must continue to burn fuel. Taking that fuel with you requires energy, which means more fuel. We cannot build a chemical powered rocket than can carry the amount of fuel needed, or provide enough thrust to get that much fuel off the ground.

Just to note, the Saturn IV that took us to the moon weighted about 5, 000, 000 pounds at lift off and had about 6, 000, 000 pounds of thrust. Of that take off weight, about 100,000 pounds was vehicle, and 4, 900, 000 pounds was propellant. That is a lot of fuel. By the time they reached escape velocity heading towards the moon, they had consumed (this is a guess) more than 95 percent of the fuel. That took something like 10 minutes of thrusting to get to orbit and another 10 minutes of thrusting (at a much lower fuel consumption rate) to get to escape velocity.

Until we find a method of propulsion better than throwing mass out the ass, we will never get anything very far out of our solar system in less than thousands of years. (BTW: I coined the term “mass out the ass” almost two decades ago.)

However, if you get a satellite into orbit, there is hope for a low speed exit. Its called a photonic laser thruster. In essence, first put one of these laser rockets in orbit. (no one has done it yet) Use solar panels to power the laser. Turn on the laser like you would a rocket engine to launch photons out the back. The photons have energy and effective mass and can provide a small amount of force, i.e. propulsion. The sun provides the continuing energy.

You might have enough thrust to increase your velocity by, let’s say, 1 foot per second each orbit. Which means that your orbit will be just a little bit higher each orbit. But you will be exchanging velocity for altitude. Each orbit you will go higher, but a little slower. Note that you will gain altitude faster than you will lose velocity. But by getting further and further away, the escape velocity drops. In some amount of time, your velocity and the escape velocity at your current altitude will match. Further thrusting will cause to you to leave the earth’s orbit.

 

[rewebster]

No, he's not. The definition in his original post was clearly for "escape velocity".

from his posts #3, #5, #8, #11 , I'm not so sure-----sure, that (the OP) is clearly a definition--but, he seems to be mixing the rational (and the definitions) for the two (escape and orbital)--especially in post #11:

isnt the orbital speed inversely dependant upon the arbital altitude?

and is there any relation between orbital speed and escape velocity?

 

[montoyas7940]

There was a rather silly television program a number of year ago (I forget the name but it starred Andy Griffith, for God's sake!) about a junk dealer who was going to make a fortune by going to the moon to recover material left there by the Apollo missions.

It was called Salvage 1. I loved it, but I was just a kid.

 

[D H]

and is there any relation between orbital speed and escape velocity?

For a circular orbit the relation is quite simple: escape velocity is the orbital velocity times the square root of two.

 

[jablonsky27]

from his posts #3, #5, #8, #11 , I'm not so sure-----sure, that (the OP) is clearly a definition--but, he seems to be mixing the rational (and the definitions) for the two (escape and orbital)--especially in post #11:

i think I'm quite clear on the two.. escape velocity is the initial speed you need launch an object at from any celestial body so that it is able to completely escape its gravitational field ie it doesn't fall back on to the celestial body.. the orbital speed is the speed that an object has when it is following an elliptical/circular path around the celestial body. so basically its speed tangential to its path is such that the object is able to maintain a certain(constant?) distance from the celestial body even though it is constantly falling towards the celestial body.. correct?

 

[jablonsky27]

For a circular orbit the relation is quite simple: escape velocity is the orbital velocity times the square root of two.

ok. thanks. can you tell me how i can go about deriving that?

 

[jablonsky27]

I'm sure NASA scientists/engineers work hard to find the optimal acceleration rate, but the reality is that escape velocity is an energy thing: whether you fly out at a small but constant speed or get shot out of a cannon at escape velocity, the amount of potential energy to be overcome is the same.

yes, i overlooked that part about it being about overcoming potential energy.

The difference with flying out at a slow speed is that you also need a constant 1g (whatever 1g is at your altitude) of acceleration force to keep your constant speed.

dont you always need to overcome that constant 1g?

 

[D H]

ok. thanks. can you tell me how i can go about deriving that?

http://en.wikipedia.org/wiki/Escape_velocity#Calculating_an_escape_velocity"

 

[jablonsky27]

D H, that link derives the expression for escape velocity. i know that. i want to know the derivation of the relation between escape velocity and orbital speed.

 

[D H]

Read a little closer: "compare this with equation (14) in circular motion". Circular motion is a hyperlink.

 

[jablonsky27]

ya, i went to that link. none of the equations are numbered and i see no equation which even remotely resembles the expression. anyways, i ll try to derive it myself.

 

[D H]

Hmm. Bad link. Try http://en.wikipedia.org/wiki/Circular_orbit" .

Alternatively, just derive it from the vis-viva equation,

v^2 = GM\left(\frac 2 r - \frac 1 a\right)

 

[pallidin]

Perhaps this can be made a little clearer... achieving Earth orbit is NOT dependent on speed, rather it is dependent on the continual lessening of the Earth's gravitational influence{and atmospheric bouyancy concernes) on the "projectile"
Even a well constructed balloon can reach heights of 5-10+ miles with no propellent. Then the atmosphere thins-out so much that the balloon can rise no higher.
Thus rocketry was born.

 

[M Grandin]

In addition to other problems with "start slowly" already mentioned, one important reason to reach as fast a speed as you can as quickly as you can is to get through the atmosphere as quickly as possible.

I wondered if you were serious about that or just a misconception. Concerning friction
losses by passing through atmosphere layer, the shorter time - the higher speed - the
more friction force - and the more energy due to friction to pass a given distance through atmosphere.

Maybe you just meant air pressure retarding rocket engine thrust -
if so you are of course right.

 

[rewebster]

escape velocity is defined in resnick,halliday,walker as 'the certain minimum initial speed that will cause it to move upward forever, theoretically coming to rest only at infinity. This initial speed is called the escape speed.'

the way escape speed is defined here gives me the impression that it is applicable only to projectiles which are given an initial thrust. during their entire trip after launch, they are not propelled.

if it is really the case then why do we need to worry about escape velocity for rocket launches? what stops a rocket traveling at say 2000km/h vertically upwards from the ground from leaving the Earth's gravitational field? afterall during its entire flight, it is being propelled upwards constantly.

This is where my thoughts are, on what you said were, going.

The rocket is trying to do both---achieve escape velocity and achieve orbital velocity--at the same time.

If an object could somehow be placed at that altitude where geosynchronous orbits occur, the thrust required to set it into a geosynchronous orbit (around the planet), to me, would have to be quite a bit more than the energy required to provide an escape velocity (away from the planet).

---I didn't do the math, but it just seems logical. For my thinking, some portion of the 'escape velocity' propelling force is used also to achieve the 'orbital velocity' speed required.

---------------------------

This, to me, is why those 'space ladders' as they are being tried and theorized about, as they are right now, would never work.

 

[D H]

If an object could somehow be placed at that altitude where geosynchronous orbits occur, the thrust required to set it into a geosynchronous orbit (around the planet), to me, would have to be quite a bit more than the energy required to provide an escape velocity (away from the planet).

---I didn't do the math, but it just seems logical. For my thinking, some portion of the 'escape velocity' propelling force is used also to achieve the 'orbital velocity' speed required.

---------------------------

This, to me, is why those 'space ladders' as they are being tried and theorized about, as they are right now, would never work.

You should do the math. A space elevator will be fixed with respect to the rotating Earth. This means that the inertial velocity of some object climbing the elevator is \omega_E r^2, where \omega_E is the Earth's rotation rate (360^o/day) and r is the distance between the climbing object and the center of the Earth. The virtue of a space elevator is that if an object climbs the elevator to geosynchronous altitude and then releases from the elevator it will be in a geosynchronous orbit with zero additional expenditure of energy.

 

[rewebster]

You should do the math. A space elevator will be fixed with respect to the rotating Earth. This means that the inertial velocity of some object climbing the elevator is \omega_E r^2, where \omega_E is the Earth's rotation rate (360^o/day) and r is the distance between the climbing object and the center of the Earth. The virtue of a space elevator is that if an object climbs the elevator to geosynchronous altitude and then releases from the elevator it will be in a geosynchronous orbit with zero additional expenditure of energy.

that sounds good (maybe), but, to me, it still doesn't make sense---


where is it getting all the tangential energy needed to achieve an orbital velocity?


climbing is vertical--it will have to draw the energy needed to reach the orbital velocity by pulling the 'weighted end' of the 'elevator' to a lower orbit.

Here's another analogy---if you shoot 'something' straight up (perfect 90 degrees) into the air 5000 miles (not counting even the air friction), it's not going to come down exactly to the same spot that it was launched from.

 

[D H]

where is it getting all the tangential energy needed to achieve an orbital velocity?

From the Earth's rotation. Use of a space elevator will slow the Earth's rotation. Nothing new here; every prograde space launch slows the Earth's rotation a tiny bit.

climbing is vertical--it will have to draw the energy needed to reach the orbital velocity by pulling the 'weighted end' of the 'elevator' to a lower orbit.

No. It draws the energy needed to reach orbital velocity by slowing the Earth's rotation rate.

 

[rewebster]

From the Earth's rotation. Use of a space elevator will slow the Earth's rotation. Nothing new here; every prograde space launch slows the Earth's rotation a tiny bit.


No. It draws the energy needed to reach orbital velocity by slowing the Earth's rotation rate.

I don't see that---

it would be like including (or having to include) the gravitation effects of the sun on the ladder too.

If you roll a marble out from the center of a record, it won't go in a straight line in respect to the record.

The energy for the increase in speed from a 8000 miles to 24,000 miles orbital velocity geosynchronous orbit has to come from someplace.
edit:
not quite right --speed is the same--(after re-reading what I wrote)

The energy to get it to the geosyn orbit is needed, and the problem is the energy to keep the object traveling up the string, straight up the string, and not lagging behind as the Earth turns underneath---as would 'the rocket' sent straight up would lag. Any force on the 'string' to try to keep it 'straight' will pull the 'weighted end' out of its orbit.

 

[rewebster]

re-edit: the speed is increased dramatically---the rotational duration is the same (totally different)


----
--this 'ladder' isn't a prograde launch

 

[D H]

I don't see that---

The energy for the increase in speed from a 8000 miles to 24,000 miles orbital velocity geosynchronous orbit has to come from someplace.

The climber has to provide enough energy to lift the vehicle from the Earth's surface to 29,400 km above the surface. That is a lot of energy: 51.36 megajoule/kilogram. Fortunately, most of this energy can be recovered by regenerative breaking in the climbers returning to Earth. The climber also has to gain kinetic energy. The climber's initial speed at the bottom of the elevator is 0.465 km/s and is 2.609 km/s at geosynchronous altitude. This amounts to 3.30 megajoule/kilogram. Fortunately, *all* of this energy comes from the Earth itself. The Earth will slow down slightly. Energy is conserved.

Here is more-or-less the same thing at wikipedia.
http://en.wikipedia.org/wiki/Space_elevator#Angular_momentum.2C_speed_and_cable_lean

The horizontal speed of each part of the cable increases with altitude, proportional to distance from the center of the Earth, reaching orbital velocity at geosynchronous orbit. Therefore as a payload is lifted up a space elevator, it needs to gain not only altitude but angular momentum (horizontal speed) as well.
This angular momentum is taken from the Earth's own rotation. As the climber ascends it is initially moving slightly more slowly than the cable that it moves onto (Coriolis effect) and thus the climber "drags" on the cable, carrying the cable with it very slightly to the west (and necessarily pulling the counterweight slightly to the west, shown as an offset of the counterweight in the diagram to right, slightly changing the motion of the counterweight). At a 200 km/h climb speed this generates a 1 degree lean on the lower portion of the cable. The horizontal component of the tension in the non-vertical cable applies a sideways pull on the payload, accelerating it eastward (see diagram) and this is the source of the speed that the climber needs. Conversely, the cable pulls westward on Earth's surface, insignificantly slowing the Earth, from Newton's 3rd law.

 

[D H]

this 'ladder' isn't a prograde launch

BTW, the agreed upon name for this device is "space elevator". There is a wealth of information on such devices on the web free for your finding if you use the correct term.

I never said the space elevator was a prograde launch. I said every prograde launch of a rocket has (*past tense*) slowed the Earth's rotation slightly. Think of it in terms of conservation of angular momentum. Prior to launch, the launch vehicle, upper stage, fuel, and satellite sit on the surface of the Earth. After all is said and done, the upper stage and satellite are in orbit while the launch vehicle and the burned fuel eventually return to the Earth. The Earth+launch package thus form a closed system. Angular momentum is a conserved quantity in a closed system. For a prograde orbit, the direction of the angular momentum of the upper stage and satellite remains more-or-less unchanged but the magnitude of the angular momentum vector is significantly larger. That change in angular momentum has to come from somewhere, and the only somewhere left in this closed system is the Earth.

 

[rewebster]

I still think the energy required will be too much for the 'elevator' string--from the speed going up that's suggested of the packet, the pressure on the string to attain the added orbital velocity, the weight of the string in the Earth's gravity pulling it back to earth, the relation to the 'drag' effect in the Earth's atmosphere and that drag added in for the ascending packet, the tension pressure at and near the 'weighted end' to counter all these, any and all things passing through the atmosphere and orbiting, etc. (how much is a 1% lean on a 144,000 km cable and how quickly can the string react?)

I still think the biggest problem will be going from about 1000 mph (on the Earth's surface) to about 6877 mph for the packet with the energy on the string to adjust for it.

http://en.wikipedia.org/wiki/Geostationary_orbit

I think the name of that counter weight should be "The Pi (in the sky)". ---(not that I'm pessimistic)

but I hope post 33 added to the OP 's question and solution

 

[D H]

I still think the energy required will be too much for the 'elevator' string--from the speed going up that's suggested of the packet, the pressure on the string to attain the added orbital velocity, the weight of the string in the Earth's gravity pulling it back to earth, the relation to the 'drag' effect in the Earth's atmosphere and that drag added in for the ascending packet, the tension pressure at and near the 'weighted end' to counter all these, any and all things passing through the atmosphere and orbiting, etc.

Calculating the static characteristics of a space elevator is just math. Plug in the physical constants, turn the crank, and out pop required capabilities. No known material came close to having the required capabilities, not even close. Unobtainium was needed. The discovery of carbon nanotubes in the early 1990s changed everything. Carbon nanotubes (short strands of them, at least) have more than the requisite tensile strength. There are still many problems, but the material has changed from unobtainium to something real.

but I hope post 33 added to the OP 's question and solution

Back to post 33 then.

If an object could somehow be placed at that altitude where geosynchronous orbits occur, the thrust required to set it into a geosynchronous orbit (around the planet), to me, would have to be quite a bit more than the energy required to provide an escape velocity (away from the planet).

Last item first: Escape velocity is a misnomer. It really should be escape speed. The direction of the velocity vector doesn't come into play when it comes to determine whether an object will escape the Earth's (or some other object's) gravity well. The only thing that matters is the magnitude of the velocity vector; i.e., speed.

An object in a circular orbit at some distance r from some object with mass M has speed given by

v^2 = \frac{GM}{r}

Escape velocity, on the other hand, is determined by

v^2 = 2\,\frac{GM}{r}

In other words, escape velocity is \surd 2 times circular orbit speed. Always. If some (small) object of mass m is located some distance r from a planet and has zero inertial velocity with respect to the planet, the energy needed to place the object into circular orbit is half of the energy needed to place the object on an escape trajectory.

 

[rewebster]

yes---

the reason I asked the OP about 'orbital' and 'escape' was that he kept referring to 'escape velocity (speed)' and 'rockets' , 'propulsion' , 'thrust' .

his question:
"why do we need to worry about escape velocity for rocket launches?"

The main purpose, usually, is to get into an orbit, and it's not a purely 'vertical' trajectory. Escape speeds (velocities) are there when needed to go beyond an Earth orbit.

 

[D H]

The main purpose, usually, is to get into an orbit, and it's not a purely 'vertical' trajectory. Escape speeds (velocities) are there when needed to go beyond an Earth orbit.

To be honest, the OP has far fewer misconceptions and misperceptions than do you. jablonsky consistently talking about escape velocity at first. He didn't bring orbital velocity into play until post #11, and then only in the sense of asking if there is a relation between orbital velocity and escape velocity. In other words, he knows escape velocity and orbital speed are distinct concepts.

I have already addressed some of your misconceptions. Now, what is this purely vertical trajectory stuff? You seem to think escape velocity must be directed away from the planet. (The same misconception appears in post 33). The magnitude of the velocity vector is all that matters. Whether the velocity is directed radially or transversely is irrelevant.

 

[rewebster]

To be honest, the OP has far fewer misconceptions and misperceptions than do you. jablonsky consistently talking about escape velocity at first. He didn't bring orbital velocity into play until post #11, and then only in the sense of asking if there is a relation between orbital velocity and escape velocity. In other words, he knows escape velocity and orbital speed are distinct concepts.

I have already addressed some of your misconceptions. Now, what is this purely vertical trajectory stuff? You seem to think escape velocity must be directed away from the planet. (The same misconception appears in post 33). The magnitude of the velocity vector is all that matters. Whether the velocity is directed radially or transversely is irrelevant.

that was in HIS first post----as was the term 'upward'---

I was trying to find out what his level was and try respond to it in post 33.


his posts:
"Why can't a rocket leave Earth at a constant speed(which is less than the escape speed) and continuous thrust and still be able to reach moon?"

"so i guess the speed and acceleeartion that the rocket has is for optimisation of fuel consumption"



Isn't radially or transverse or vertically "away from the planet"----If you're implying that going into the planet is possible for an "escape velocity" (that term was also used before I entered the thread)--I guess you can use enough thrust to do that too.

Re-reading the entire thread may help.

 

[D H]

Forget about jablonksy. I am worried about your level of understanding.

Transverse means crosswise; i.e., not radial. Escape velocity is a bit of a misnomer. Escape speed would be a much better term. All that is needed to determine whether or not some tiny object will escape the gravity well of some massive point mass object is the distance between the object and the point mass and the magnitude of the tiny object's velocity vector with respect to the massive object. The direction of the velocity vector is irrelevant.

A non-point mass massive object (e.g. the Earth) does present an additional complication of course. It's kind of hard to escape a gravity well when you crash into it. Ignore that complication.

Suppose a vehicle planned for some remote mission has been placed in a circular orbit around Earth. It needs to achieve escape velocity to begin its journey. How should it fire its thrusters? The answer is, along the velocity vector. Since escape velocity is \surd 2 circular orbit speed, a 41.4% increase in the orbital velocity will give the requisite escape speed. Achieving escape velocity by thrusting radially would require a lot more fuel, more than twice as much more fuel.

 

[rewebster]

ahhhhh--that's nice---but you don't have to worry about me---


I want to worry about you!-----


(I have no problem with the rest of your post---and jablonksy should link the definitions in post #23)

 

[jablonsky27]

I have no problem with the rest of your post---and jablonksy should link the definitions in post #23

those weren't definitions i picked up from someplace.. i wrote em myself.. trying to show you that i know that they re two different things..

 

[tanditchener]

To clarify, 'escape velocity' simply means: the mininimum vertical velocity of an object directly after an impulse such that the Earth's gravitation is not capable of attracting the object back towards itself. At sea level, for instance, a football would need a speed of 11.2 km/s directly after you kicked it in order for it to escape into space.

 

[D H]

To clarify, 'escape velocity' simply means: the mininimum vertical velocity of an object directly after an impulse such that the Earth's gravitation is not capable of attracting the object back towards itself.

The vertical velocity part is wrong. Escape velocity is a bit of a misnomer. It would be to call it escape speed because the only thing that matters with regard to whether some object will escape or not is the magnitude of the velocity vector. The direction in the velocity vector is pointing is completely irrelevant (so long as the object doesn't crash into the planet, that is).