# Homework Help: Ultra long range projectile problem

1. Mar 11, 2009

### putongren

1. The problem statement, all variables and given/known data
Physics students know that if one fires a cannon ball and one assume that the earth is flat and the gravitational strength is constant for all height, we use the equation Range = (Vo2/g)* sin 2 θ, Vo = initial velocity, g = 9.8 m/s2 , θ = projection angle. How do we find the equation for range if we do not assume the earth is flat and gravitational strength is constant for all height, neglecting air resistance? Like for ultra long range projectiles.
2. Relevant equations

3. The attempt at a solution
My attempt to solve the problem is to use polar coordinates instead of Cartesian coordinates. The origin is centered on the center of the earth and the y coordinate axis intersects the zenith of the projectile motion. I do not know if this is correct, but I assumed that the motion of the projectile follows a parabolic flight path. So the equation for parabolas in polar coordinates is r = { l \ {1 + e *cosθ} }, l = semi-latus, e = eccentricity. But after that, I have no idea how to do the problem.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 11, 2009

### Staff: Mentor

That's a great idea! I wouldn't have thought of that right away, but what a natural approach.

Just solve for the landing point when the distance to the origin is the same R as the firing point.

3. Mar 11, 2009

### Astronuc

Staff Emeritus
While g is constant, what is the effect of the projectile's tangential velocity?

What happens if the projectile is fired in the direction of rotation of the earth? How about in the direction opposite the rotation of the earth?

4. Mar 11, 2009

### putongren

wait, my math sux, I dun know how to do that.

5. Mar 11, 2009

### putongren

I forgot to mention to neglect rotation of the earth.

6. Mar 11, 2009

### Astronuc

Staff Emeritus
Ah - so then one only needs to consider the intersection of a circle with a family of parabolas, which have the axis parallel to the diameter of the circle.

7. Mar 11, 2009

### putongren

Correct, to clarify it further consider only 2 dimensions (the earth is a circle not a sphere).

8. Mar 11, 2009

### Staff: Mentor

Cool. Show us your work and the solution!

9. Mar 11, 2009

### putongren

I think there's some confusion. Do NOT assume that gravitational strength is constant for all height. That should make the problem a little harder.

10. Mar 11, 2009

### Staff: Mentor

A little, but not much. Show us the relevant equations, and your attempt at a solution. This is homework/coursework, so per the PF "Rules" link at the top of the page, we are not going to do your work for you.

Show us your work.

11. Mar 11, 2009

### putongren

Ok here's my try.

Use polar coordinates. Superimpose r = radius of earth and r = { l \ {1 + e *cosθ} } (if you plot the graph of both equations together, you will see an ellipse on top of a circle with them centered on the same point). Then you can solve for θ (There are two answers, θ1 and θ2 ). Once you know θ1 and θ2, you know the intersection of the flight path of the cannon ball with the surface of the planet. The arc of the circle that connects the points where the circle and ellipse meet is the range. Range = 2*diameter*π2/(θ12).

The question remains is how to determine l = semilatus rectum and e = eccentricity. These two values determine the shape of the parabola, and hence the flight path of the cannon ball. I know that these two values is somehow connected with the equation of gravity, which is F = GMm/r2. That's all I know how to do.

12. Mar 12, 2009

### putongren

I forgot to mention something. Make sure initial velocity of the cannon ball and angle at which the cannon ball is shot at is in the equation to determine the range.

13. Mar 12, 2009

### Astronuc

Staff Emeritus
There are two angles to consider. One is the angle swept by the arc of the range. Instead of being a flat surface, the range is a circular arc, and the length is Rθ, where R is the radius of the earth's surface, and θ is the angle subtended by the arc.

Then there is the angle φ of initial projectile velocity with respect to the earth's radius or the tangent to the surface.

There is also a constraint imposed by selecting a parabola (as opposed to some other form) to intersect with a circle.