# Force needed to reach Lagrange Points

1. Apr 18, 2010

### curious111

I'm trying to figure out how much force, over what period of time, is necessary to reach an earth-moon Lagrange point. L1 is about 323110 kilometers from earth, and an object there could remain (more or less) stationary relative to the earth and the moon.

Earth gravity is working against the object the entire time it is traveling, until it gets to L1. And, of course, you don't want to have to apply a force the entire time the object is traveling. So, is the answer that the force needed is however much force you need to reach escape velocity? And then you would carry that speed until some other force slowed or stopped the object? What if you want to accelerate to a given speed such that by the time the ever-decreasing force of Earth's gravity brings the object to a stop you have reached L1?

I may not know how to solve them, but I am sure there are equations answering these questions...

Well, this is not a homework question. I am well out of college and didn't study math, so I am here because I don't have the basic understanding to do this myself. To the extent you can help me figure out what to do, how to do it, and what the answer is, I would appreciate it. Thanks.

Edit: I didn't give you the payload mass. So I guess that's the other problem I have is that in order to solve this you need to know a lot about your rocket. But I guess to pick or design a rocket you'd have to know how much force you needed from your rocket to move the payload? So let's assume you're carrying a 10 kilogram object, just to make things easy.

Last edited: Apr 18, 2010
2. Apr 18, 2010

Force is not a good concept here. Energy is much better. Earth has a -1/r gravitational potential, the moon too. You find the total potential energy of the object at the Lagrange point and the on the surface of the earth, end then you know how much kinetic energy you need to get there (if there was no atmosphere). This is less then the escape velocity, but you would probably use more speed then what you calculated, because the object gets slower and slower on its way to the Lagrange point.

Last edited: Apr 18, 2010
3. Apr 18, 2010

### D H

Staff Emeritus
Close. You forgot to include the kinetic energy of the object once it becomes stationary with respect to the Lagrange point.

Addendum You could of course work in a frame that is rotating at the Moon's orbital rate. The object is stationary in that frame. However, rotating frames are probably a bit over the top for a beginner question. Nonetheless, that is the preferred frame for working out physics in the vicinity of L1.

Last edited: Apr 18, 2010
4. Apr 18, 2010

### curious111

thanks for the replies, I appreciate it. DH, you are right that is a little over my head. Could we work through it, and then see what the results mean in a practical application?

5. Apr 18, 2010

### D H

Staff Emeritus
Well, before giving too much help, and the wrong help, two questions:

1. Is this homework? Be honest, please.
2. What is your education level? Once again, be honest.

6. Apr 18, 2010

### curious111

as i said in the first post, no. I am just interested in this stuff and let my mind wander, but don't have the math/physics background to answer my own questions. I gather cheating on homework is a common problem here, so that's why I mentioned that it wasn't homework.

I have a JD from a T14 law school, i.e., I'm highly educated, just not in math. :) My undergrad math clases were in formal logic, which was invaluable on the LSAT but useless in this context.

i do appreciate the help!

7. Apr 18, 2010

### curious111

These are a few things I found on my own:

Rocket equation: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
Delta-V calculator: http://www.talisiorder.ca/worldbuilding/rocketequation.php [Broken]
Target velocities:

So, should I really be asking how much force is needed to reach 12 or 13,000 m/s?1

I hope my question is clear...
___

1 But it would be less than that, right, because once you reach L1 moon gravity begins to cancel out earth gravity?

Last edited by a moderator: May 4, 2017
8. Apr 18, 2010

### D H

Staff Emeritus
Delta-v from LEO to the Earth-Moon L1 point is about 3,800 m/s.

About the Earth-Moon L1 point. This is *not* the point where the Moon's gravity cancels that of the Earth. If the Moon was not there, an object at the L1 point would orbit the Earth in less than 27.32 days. The Moon is there of course; it's gravity reduces, but does not cancel, the Earth's gravitational influence. The reduction is just that needed to make the object orbit the Earth in 27.32 days.

Force is a bit of a red herring here. The ideal rocket equation that you found is in expressed in terms of delta-v, not force.

9. Apr 19, 2010

### spacester

Hi curious111,

Force is the wrong concept here because of the way that you actually make the flight happen. With orbital transfers, you do not hold down the gas pedal as you travel. Instead, you burn your rocket engines in order to change the shape of your elliptical orbit. If you do it just right, your new orbit will cross the orbit of the place you are trying to rendezvous with. So you burn the engines for a short period and then coast. When you arrive, you again fire your engines, this time to match orbital paths.

I would be happy to post an equation or two even a lawyer can handle :-) if you are interested.

10. Apr 19, 2010

### spacester

Actually, I'll try to explain without equations first.

Once you get to orbit, it is not that hard to calculate the requirements to transfer from one orbit to another. What you can do is just find the difference in the specific energies of the two orbits. Once you know what thing you are orbiting (Earth in this case), the specific energy of an orbit is only dependent on the equivalent circular radius of the orbit, IOW the altitude.

Specific Energy is the energy of the orbit divided by the mass of the thing orbiting (your 10 kg). We use specific energy because then we can talk about the subject of orbital transfers without worrying about the mass of the thing transferring. That's the whole thing about delta-v as well: the number applies regardless of what spacecraft mass you are talking about.

The orbital energy is an interesting concept (and I should note that this method of mine is unorthodox and some smart people are going to think I am cheating, if things go the way they usually do - I don't post here much). It is the BOUND energy, representing the energy of the motion of the object in a gravity field (where the numbers give you an elliptical rather than parabolic or hyperbolic orbit). It is actually less the further away you are. A bound energy of nearly zero is what an object just barely within the gravity field has.

On a related note, be aware that 'escape energy' is of virtually no use in these calculations. That term is very mis-used by the, er, curious. Escape energy is that specific energy required to take an object from its current location in the gravity field (usually from the surface of the planet) and apply it all at once and have the thing coast straight up such that it eventually stops and does not return. This has little to no relation to something that would happen in real life, especially since you have to ignore other gravity fields, which is ludicrous in the real solar system. It is called 'escape' because any less energy and you fall back to the surface. It is a (bad) measure of gravity field strength.

So once I give you the equation, it will be simple (for those used to plugging numbers into an equation and turning the crank correctly): You find the radius of the starting orbit, which is actually the same thing as the "semi-major" axis of a real-life elliptical orbit. You plug that in and get the starting orbit's bound energy. Then you use the L1 altitude to find the bound energy of the second orbit. Take the difference, and you have the energy change. From there you use the equation for Kinetic energy to convert to delta-V. Real-life "gravity losses" are ignored here.

This delta-V number does not tell you how much dV is needed for the first burn vs the second burn, it is just the total. You need to solve for the parameters of the transfer orbit to get the allocation between the two rocket engine burns.

Once you get the dV, and the mass, you actually can get the force required. First you have to specifying the time spent burning engines. It is a very simple equation.

11. Apr 19, 2010

### D H

Staff Emeritus
That is not the reason force is the wrong concept here. It is not that force is the wrong concept. Ultimately, force is a very essential concept; force is after all what ultimately makes the rocket go. The problem is that force is not a central concept. Force typically comes into play near the end of the calculations.

Moreover, the two burn transfer, aka burn-coast-burn, is not how all orbit transfers are performed. Launch, for example, is a burn, baby, burn process rather than a burn-coast-burn process. SMART1 used a burn, baby, burn process; it's puny thruster generated a very small amount of thrust. SMART1's thruster was for about half the time during the first 13 months as it gradually (very gradually) transferred from low Earth orbit to the Earth-Moon L1 point.

Anyhow, curious111, look at these webpages, http://exploration.grc.nasa.gov/education/rocket/guided.htm. The target audience is an intelligent high school student who is on a math and science track. That's about the right level for most lawyers I've run into who have an interest in "rocket science".

12. Apr 19, 2010

### spacester

Hi D H,

I am not here to argue or get into a pissing contest or quibble over non-essential points or explain Lambert's method before discussing a Hohmann transfer :-) let alone the wonderfully complex parametric equations used to solve constant thrust trajectories.

I'm just here to help out once in a while. I am actually a fan of yours, and I've studied orbital mechanics a lot and want to share some of it. I am hoping there is an increased interest in the topic as we start thinking about getting to and from NEOs.

curious111's question was way too juicy for me to ignore. I love explaining the topic to smart people who lack the background we do.

And hey, you gotta admit that your first instinct when you saw the thread title was that the concept of delta-V needed to supplant the reference to force, right?

13. Apr 19, 2010

### D H

Staff Emeritus
Those first two are a bit important. They might help curious111 a bit. The last one is over-the-top for someone new to the general topic, and most likely new to the math. So, without further ado,

Transfer orbit
Suppose you want to make a spacecraft that is in some orbit move to a different orbit. Suppose the spacecraft's engines have a good amount of oomph (not all do; SMART-1's single engine had oomph). What is done is to burn the engines to change the vehicle's orbit. Do this right (and there are lots of ways to do this right) and the new orbit will intersect the desired orbit. This intermediate orbit that takes the vehicle from the initial orbit to the desired orbit is the transfer orbit. The vehicle just coasts along this transfer orbit. When it gets to the desired orbit, it first its engines again to change the orbit once again, this time changing from the transfer orbit to the desired orbit.

Hohmann transfer
As mentioned, there are lots of ways to accomplish this transfer. Some transfer orbits will intersect the desired orbit twice, others just once; the transfer and desired orbits are tangent at the intersection point. If the vehicle did not fire its engines the second time, the vehicle will simply continue on the transfer orbit. This will eventually take the vehicle back to the starting point, which was a point on the original orbit. The transfer orbit will also intersect the original orbit one or two times. One time means that the transfer orbit is tangent to the original orbit. Making the transfer orbit tangent to both the original and desired orbits minimizes the amount of energy used. This can be a bit tricky to calculate in general. It's not so tricky when both the original and desired orbits are circular orbits in the same orbital plane. This is a Hohmann transfer. High school students (smart high school students) can calculate a Hohmann transfer.

Lambert targeting
It would do little good to transfer a satellite from low Earth orbit to geosynchronous Earth orbit and have the vehicle end up in a geosynchronous orbit over India when it is supposed to be over the Western Hemisphere. It would do little good to transfer a vehicle that is to dock with the ISS from orbit insertion to the ISS' orbit, but only have the vehicle end up tens of thousands of kilometers behind the ISS. In general a vehicle not only needs to get to the desired orbit, it needs to hit a specific point on the desired orbit at a specific time. This is what Lambert targeting does. It ensures the transfer will take the vehicle to a specific point in space and time. Doing so might not be optimal in terms of fuel usage, but such is life.

A couple of sports analogies:
• Football (America and Canada). The quarterback is throwing the ball to a moving target. He needs to throw the ball to where the receiver will be, not where the receiver is. The ball's and receiver's trajectories must intersect in space and in time.
• Football (everywhere else): The same concept applies when a midfielder makes a diagonal pass to a striker. The striker is a moving target. That the ball's and the striker's trajectory intersect at some point on the pitch is not enough. If the ball has too much pace the striker won't be able to receive it; too little pace and the ball will end up behind the striker. Those trajectories have to intersect in space and in time before the fans cheer "nice ball!"

14. Apr 19, 2010

### spacester

D H,
You're not much into dialog are you?

Do you have a proprietary interest here I should know about? Because you seem to me to be acting like I am a threat by way of my simple participation.

15. Apr 19, 2010

### spacester

Oh and very nice write-up as usual.

16. Apr 19, 2010

### D H

Staff Emeritus
My PhysicsForums paycheck (\$0.00) does not do much to keep the wolves at bay. I've got some very real deadlines in my real job due [strike]today, tomorrow, Thursday, and Friday[/strike] Merde! Every single day of the week! --- and I want to take Friday off. I need to keep the dialog short. Sorry if that comes off as being brunt. You have raised some good points.

17. Apr 19, 2010

### spacester

Cool. I really like the football analogy. I'll go back to lurking for the most part. The subject I'd like to discuss is too off topic.

18. Apr 20, 2010

### D H

Staff Emeritus
There is a simple solution to that problem: Start a new thread if there is some burning issue you would really like to discuss but is off-topic to the topic at hand.

curious111, we haven't heard from you as of late. Are you still around?

19. Apr 20, 2010

### Magellan1

OK, so here are my questions:

Leaving the ground:
1. What is the question that needs to be asked to find out how much fuel/force/speed is necessary to reach LEO? I feel like right now we know the delta-V needed. And we want to get to LEO reasonably quickly. So it must require a significant force applied over a relatively short period of time? Or, to make sure I'm not conflating terms, how much thrust is needed to reach that velocity? And how do you determine what that means, and how much energy/fuel is needed to produce that amount of thrust?

2. Once you find out what's needed to get off the ground, how do you aim? I've seen graphs of rocket trajectories. How do you make sure it doesn't fall back to the ground once it reaches its apogee?

Orbit:

Once in LEO, how do you "find" or express what your orbit is? Then how do you find out how much additional change in velocity you need to make your orbit "broader" / "wider"?

And, I think that raises the first question again, how do you find out how much thrust or force is necessary to achieve that delta-V?

I hope I'm being clear. I really appreciate you guys all humoring me. Thanks much

20. Apr 21, 2010

### D H

Staff Emeritus
Obviously force is a concern here. A rocket that generates less upward force than the downward force of gravity is not going anywhere; it won't even be able to take off. The force generated by the thrust must counteract gravity, and then some.

In terms of final versus initial conditions, a launch vehicle has to
• Lift the vehicle vertically by 185 km or so (orbit insertion altitude).
• Accelerate the vehicle (horizontally) to 7.8 km/sec or so (orbital velocity).
• Do this rather quickly (9 minutes or so).

Rockets that have the oomph to do that are big beasts. They have to be launched vertically. By the time the vehicle has left the atmosphere and is nearly on orbit, the rocket has to be firing nearly horizontally. Between launch and orbit insertion the vehicle needs to pitch down 90 degrees.

Beyond this, you are (innocently) asking a bit too much. You are asking us to write a book. Several books, in fact. That said,
• Open courseware slides for "Space System Design" are at http://www.princeton.edu/~stengel/MAE342Lectures.html.
• A big (and arguably overly bureaucratic; I suspect Elon Musk would a bit askance at this) book on the Launch Vehicle Design Process is at http://trs.nis.nasa.gov/archive/00000569/01/tp210992.pdf [Broken].

At a minimum the vehicle needs to do a 90 degree pitch down between launch and orbit insertion. The best way to do this is via a gravity turn. Ideally, a launch vehicle will have a zero angle of attack throughout its flight. In practice, this isn't possible because the vehicle has to be launched vertically. A tradeoff is involved here, and max Q (google that phrase) plays a big part in that tradeoff.

Simple: You make sure the vehicle is moving at about 7.8 km/second (horizontally) when it reaches 185 km altitude.