# Angular momentum at an angle!

1. Mar 27, 2004

### pantalaimon

This sum is really wrecking my head.

a satellite orbits the earth in a circular orbit, a height 'b' above the earth's surface. A missile is launched from the earths surface(in the satellites plane) with a speed V at an inclination theta to the vertical, and h= RVsin(theta) for the missile (R being the radius of the earth), we want the missile to hit the satellit at the apogee of the missiles elliptical orbit.

The question wants me to prove that some massive equation holds, i amn't worried about that, what i AM really annoyed about is the vagueness of line 3 of the above paragraph.

Surely why would my professer ask us to set up a vertical axis, when we can just have the velocity be 100% tangential and then we can assume perigee to be that position.

Obviously i am reading it wrong, i don't want the answer to the problem, i just want to know what are they alluding to when they say that the missile is launched with a speed V AT AN ANGLE THETA TO THE VERTICAL? If the angular momentum is given by RVSin(theta) then our ellipse will NOT be working with the condition that the velocity at perigee, but VCos(theta). Am I wrong to assume this. I would appreciate it if anybody could clear up this mess.

thanks.

2. Mar 27, 2004

### NateTG

Got to agree with you on the poor phrasing. I expect that they mean with an angle &theta; from the normal vector to the earth's surface.

3. Mar 27, 2004

### pantalaimon

Yeah, but do we proceed to determine the elliptical equation wrt the direction of V, as this would make perigee the starting point, I wish i could show you's a drawing that i drew. Anyway, thanks for agreeing with the fact that i amn't a complete paranoid freak abut the wording.

4. Mar 27, 2004

### enigma

Staff Emeritus
h is a constant for every orbit, and is equal to R x V, RVsin(theta), or RVcos(flightpathangle). We know that theta is defined as inclination from vertical, or inclination from R. We also know that at apogee and perigee, R and V are normal to each other. Does that help any?

5. Mar 27, 2004

### pantalaimon

I know we're on the wavelength so, just let me clear something up. theta is then NOT a allowed to be anything else but 90 degrees, at launch as V must have no r' componant, i.e all of the velocity is purely rotational. So let me rephrase;

We shoot a missile at an angle theta to the vector r, where r points out from the centre of the earth.

so in polars;

r(theta) @ theta=0 is nothing other than perigee.

and the angular momentum as per the question is RVsin(theta).

okay grand, i understand this and also that it is conserved throughout its journey.

BUT, if theta not=90 then r is not perpendicular V, and this implies that there IS a rotational part of the velocity which implies that that point is NOT perigee. Do you agree that it is impossible for that point to be perigee. Unless of course theta=90.

Am i wrong, please tell me that i am!

6. Mar 27, 2004

### enigma

Staff Emeritus
No, you're not wrong, but...

You don't have to be at perigee when you take off (and in reality, you never are). If you were at perigee, you'd be doing a Hohmann transfer problem, and taking off like an airplane.

7. Mar 27, 2004

### enigma

Staff Emeritus
To elaborate a bit more,

if the Earth weren't there (but the gravity remained), the missile would be on an orbit which would have perigee inside the Earth's radius.

Since the Earth is there, the missile is only doing a suborbital hop. There isn't any reason to give it an orbital velocity when all it needs to do is get up to the correct altitude with near zero tangential velocity to blow up the satellite.

Welcome to the forums, BTW...

8. Mar 27, 2004

### pantalaimon

Thats exactly what i was thinkin. But does that mean now that the earth will not be at the focus of the ellipse. And further more it will not obey the elliptical equation that we all know:

R(theta)=L/(1+ecos(theta))?

Making a big mess of everything?

9. Mar 27, 2004

### enigma

Staff Emeritus
No, the Earth will still be at the focus. You'll have a pretty high eccentricity, so perigee will be really low... I'd guess somewhere between a few hundred to 1-2 thousand km.

Imagine you can go through the Earth, and fly the missile backwards in your mind. You're not taking off with a FPA of 90 degrees, so you'll have some tangential velocity. As you lose "altitude", you'll loop around the center of the Earth reeeealy quickly, and shoot up the other side.

I don't think you even need to take true anomaly into account at all...

10. Mar 27, 2004

### pantalaimon

So you're sayin perigee WON'T be inside the earth?

11. Mar 27, 2004

### enigma

Staff Emeritus
Sorry,

My guesstimate was in km from the center of the Earth, not altitude.

12. Mar 27, 2004

### pantalaimon

cos if perigee is the point of closest approach and it was launched from the SURFACE, surely the only closer you canget is actually INSIDE the earth.

I amn't sayin you're wrong, but logically speaking;
If the missile is launched from the surface of the earth, and the velocity is NOT purely rotational. Then that point is not perigee, therefore there is a closer point to the centre of the earth. Thus the equation of true anomoly is kinda worthless cos we don't know where the missile is at launch in polars. like i CANNOT say that

r(0)= l//(1+e)

As you would normally in such a problem, you know.

13. Mar 27, 2004

### pantalaimon

grand, i was gettin worried

14. Mar 27, 2004

### pantalaimon

So am i safe to say that perigee will be a little to the left or to the right of the initial launch position.

15. Mar 27, 2004

### enigma

Staff Emeritus
You don't really need to worry about perigee, do you?

Depending on the actual launch conditions, it could be almost anywhere so long as it is inside the surface of the Earth, and it isn't 180 degrees away.

16. Mar 27, 2004

### pantalaimon

Well yeah. Cos i need to write the angular mom in terms of V, sin(theta), R, and b. also in terms of little g(gravity) i don't know how i am gonna bring that into it.

But i know that i'm gonna need the energy equation

1/2(V)squared-GM/R = GM/2l((e)squared-1)

I will get rid of the e's and l's using the ellipse equation i mentioned earlier, but i need to know what r(0) is and r(180) is.

17. Mar 27, 2004

### pantalaimon

A satellite is circiling the earth at a height 'b' above the earths surface. A missile is fired from a launching site on earth, in the plane of the satelite's obit with speed V at an inclination theta to the vertical, so that

h=RVsin(theta)

for the missile where R is the Earth's radius. If the missile and satellite collide at the apogee of the missile's path show that;

(RVsin(theta))^2 = (R+b)[(R+b)V^2 - 2gRb]

That's the question word for word, i still believe the question to be lacking in clarity.

18. Mar 27, 2004

### enigma

Staff Emeritus
I don't think you don't need perigee.

The semilatus rectum, p (I think you're referring to that as 'L'), is defined as:

$$p=\frac{h^2}{\mu}$$

$h^2$ is your left hand side of the equation.

That should get you started... just push variables around till it clicks.

Last edited: Mar 27, 2004
19. Mar 27, 2004

### pantalaimon

Yeah i have that formula. Thanks for your time though. I'm gonna keep trying.

20. Mar 27, 2004

### pantalaimon

To be honest i thought i'd be waiting a week for a response.