# Need help with rocket problem

1. Oct 13, 2006

### Noone1982

Hello,

I am having an awfully hard time working out this rocket problem. Say you want to have a 2D representation of a rocket launching from the earth, either it will crash right away, have a decaying orbit or a stable orbit.

If you launch a rocket at, say, 88 degrees it will have a velocity in the r-hat and phi-hat direction. It will continue to go higher and faster as the rocket consumes its fuel.

To plot this using cylindrical coordinates, we need to know phi as a function of time. But how can one do this? The only way Iv thought how to do this is calculate the angle phi incrementally examining the triangles from one position to the next.

Perhaps there is a better idea?

2. Oct 14, 2006

### Noone1982

Perhaps let me restate this,

The velocity of the rocket is given as,

$V\; =\; V_{ex}\ln \left( \frac{Mo}{Mo-KT} \right)-g\left( h \right)t$

Where Vex is the exhaust speed, Mo is the initial speed of the rocket, K is the burn rate, t is time, g is gravity

Now if we launch the rocket at a slight angle, it will have velocity in the r-hat and phi-hat direction. Now the height is not simply going straight up, but a curve around the earth and the velocity is dependent on this, we need to know the angle phi as a function of time.

I suspect I am making this more complicated than need be, I'm pretty good at that ;)

Here is an image that could help:

[Broken]

Last edited by a moderator: Apr 22, 2017 at 1:25 PM
3. Oct 14, 2006

### Sojourner01

I believe what may complicate this solution is whether the force from the rocket's engine changes direction relative to the radius of the orbit. This would make the acceleration in each dimension a function of its position, which sound horrendous.

Consider what you know. You know that at a given height, or coordinate r, there is a certain speed required to keep the rocket in a stable orbit at that height. The force exerted by the rocket's engine has a radial component Fr and a tangential component F(phi). As long as these remain constant, this question should be straightforward. If you're assuming you want to set the orbit to where the rocket runs out of fuel, solve a projectile equation in the dimension r and a circular motion equation in dimension phi for equal t, then that should get you the force components and thus the ratio between them can be used to determine the angle of launch.

At least, that's what I can see in overview. I haven't attempted a problem of this type before.

4. Oct 14, 2006

### Noone1982

That's right, the rocket's engine changes its orientation continually with the acceleration as a function of position. I never considered this. Egads! That is horrendous!

Considering this fact, I'm not sure if I can proceed with this right now. I am making a program that graphs a rocket as a function of time. My advisor asked me to up the ante and show a 2D representation of it obtaining an orbit. The acceleration would have to be considered as a position of time for an accurate model.

I'm not that worried about obtaining what angle to launch the rocket at the moment. I need a general model.

Lets see what we know,

Rocket Velocity = v(t) - g(h)t

Now phi = phi(?)

I'm not sure what to do about phi. Without knowing phi as a function of velocity, time or height the program cant graph in two dimensions.

I was experimenting with some relationships with the green triangle above, but I'm not sure if that's a valid way to do it.

5. Oct 14, 2006

### OlderDan

As a first effort, you could do a straight up rocket. Your equation assumes the thrust is straight up away from the center of the earth. Since you start off with some horizontal velocity do to the earths rotation, you could still achieve orbit. Tilting the rocket could be added later. You would probably want to start by assuming a launch at the equator.

6. Oct 14, 2006

### Noone1982

That is what my original model is doing - a straight up launch. Though delaying the tilt still doesn't solve the problem of phi(t) a(position), etc

Two tests, two homeworks, one lab report and an orbit simulator due by thursday. This week will be no picnic.

7. Oct 15, 2006

### Sojourner01

If your launch is straight up, your angular momentum is constant - the amount of angular momentum you had when you were on the earth's surface, rotating around its centre of mass. Therefore, if you're changing only your altitude, the parameters of the resulting orbit can be found by rearranging:

angular momentum at launch position = angular momentum at orbital height

where orbital height = climb rate of rocket x time of rocket burn
and launch position = radius of the earth (obviously)

8. Oct 15, 2006

### Noone1982

Yes, my straight-up model uses what you just said, Sojourner. It works perfectly - it was my advisor's idea to make a 2D representation of it obtaining orbit.

9. Oct 15, 2006

### OlderDan

I'm having trouble figuring out where your problem lies. Your original post said you were having trouble finding phi as a function of time. Are you now saying you know phi as a function of r for the straight up rocket? Doesn't your rocket equation work tell you r as a function of t?

10. Oct 15, 2006

### Noone1982

My original model which I have working now doesn't even use phi, just the radial component. I'm trying to up the ante and make a 2D representation of it obtaining orbit, with that I would need to graph both the radial and phi component.

I know r as a function of time, that's easy.
I do not know phi as a function of time.

Making matters worse, my original model just assumed the rocket's orientation never changed. The cone is up, the nozzle is pointed down - that's easy! However, Sojourner pointed out that I mistakenly forgot that the orientation changes with respect to position(phi,r) The rocket would initially have almost all of the thrust in the r-hat direction but would eventually have the thrust all in the phi-hat direction as the rocket obtained an orbit.

I'm not sure how to account for how the changing orientation with respect to position affects acceleration and the pesky problem on how to define phi as a function of time.

11. Oct 15, 2006

### OlderDan

The rocket does not have to change orientation to achieve orbit. If the earth were not spinning, you would have to tilt the rocket to give it a tangential velocity component, but your rocket already has tangential velocity at lift off, and it will retain tangential velocity as long as all forces acting are radial forces. The only question is how much tangential velocity will it have as the rocket rises.

Fortunately, as long as the rocket thrust is vertical, the question has a simple answer. As stated earlier
The rocket will have the same angular momentum all during the trip. The fuel that is expended also experiences only a radial force, so its angular mometum will also be conserved. Angular momentum is just the tangential component of velocity times mass times r. The gas leaving will not affect the anglular momentum of the rocket, so you don't have to worry about the angular momentum of fuel being carried by the rocket. Focus on the rocket body, which has constant mass, and determine the tangential velocity as a function of r. That is what you need to plot the path of the rocket.

Once the fuel is expended, the rocket will have velocity with radial and tangential components, but from that point on it will have constant energy and angular momentum. It will be in orbit. The shape of the orbit will depend on the energy and angular momentum. If it has enough of both, the orbit will never bring the rocket in contact with the earth again.

12. Oct 15, 2006

### Sojourner01

I think Noone's saying that he's already worked through the vertical model, Dan.

The next step is to create a model where you assume the rocket's pitch relative to the radius is constant. From this assumption, the tangential component is constant and the rocket gains a fixed quantity of angular momentum per second.

So, dL/dt = constant, and equal to F(thrust) cos (launch angle)

13. Oct 15, 2006

### OlderDan

I thought so too, but if you read his post #10 he is saying he has not yet included anything about phi. Before delving into the issue of tilting the rocket, he should at least be able to incorporate the angular motion for a central force problem.

14. Oct 15, 2006

### Noone1982

Why would I need phi in a purely vertical model?

15. Oct 15, 2006

### OlderDan

Because vertical means directed away from the center of the earth. Before launch, your rocket is moving in a circle around the earth's center. Applying a vertical force will not alter the angular momentum it has before launch. If you applied just enough thrust to lift the rocket off the pad and hover, it would follow a circular path. There would still be a net force acting toward the center of the earth, changing the direction of its velocity but not the magnitude. This effect does not disappear when the rocket accelerates upward, but the horizontal velocity changes just enough to conserve the angular momentum.

16. Oct 16, 2006

### Noone1982

Ahh, I see what you're saying. That the rocket will obtain a geo-synchronous orbit at a specified height?

17. Oct 16, 2006

### OlderDan

It will definitely not be geo-synchronous because the angular velocity must decrease as r increases. Since it starts out with the correct angular velocity for geosynchronous orbit, it must have too little angular velocity at higher altitudes. My instinct is that you will not be able to achieve a circular orbit with a vertical thrust. The only thing you can change with a vertical thrust in your model is the fuel burn rate. That should affect the resulting orbit.

You could get around this limitation of the model by using a two stage thrust. First, let the thrust be vertical to achieve a desired maximum altitude. Use conservation of angular momentum to account for the phi motion. After the vertical burn is complete, the rocket will drift toward its maximum altitude. Then as the rocket is approaching maximum altitude let it fire in a tangential direction to give it the velocity needed to achieve circular or near circular orbit. For geosynchronous orbits, only one constant altitude is possible.

Here is a site that has the equations relating the shape of an orbit to the total energy and the angular momentum.

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/EllipticOrbits.htm