Escape velocity for rocket launches

[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).