Where is second focus?

Question

A rock thrown at 45 degrees travels along parabolic
trajectory for longest possible distance.
For example, if initial velocity is 20 m/s, the rock lands
40 meters away if thrown at 45 degrees.

According to Kepler laws, however, the trajectory of the
rock muct be ellipse, with one focus at the center of the
earth. Where is the second focus of this ellipse?

Answer 1

At Earths surface, or from whatever altitude you are throwing the rock.

Answer 2

Wow, this is a really tricky one. I hope I got my math right.

scythian_1950's answer is only correct if you throw the rock faster than 7 miles per second (!)

kennyk's answer is clearly wrong--here's why:
Scenario A: You shoot a ball from the ground and it just clips the corner of a 100-ft building.
Scenario B: You shoot a ball from the top of the 100-ft building. You adjust its initial speed & angle so that it matches the speed & angle that Ball #1 had when it clipped the corner. That means Ball #1 and Ball #2 both follow the same parabolic arc (& hence the same ellipse) from the top of the building onward.

So clearly, the focus is in the same spot for both scenarios. But kennyk says that in Scenario 1 the focus is on the ground, while in Scenario 2 the focus is at the top of the building. That's a contradiction.

Okay, now for my math. First of all, let's pretend that the earth is not spinning. The spinning earth adds a large eastward component to your launch velocity, so you have to take that into account when figuring out the ellipse. But let's worry about that later.

Here's the equation for an ellipse aligned with both foci on the y-axis, and the lower focus at the origin:

x^2/b^2 + (y - ae)^2/a^2 = 1 (eq. 1)

(where a = semimajor axis; b = semiminor axis; and e = eccentricy = sqrt(1 - b^2/a^2)). So, you can imagine this as our ellipse, with the earth's center as the origin of the cooridinate system, and the tippy-top of the ellipse poking above the earth's surface at the top of the diagram.

Now let's express this in a different coordinate system, with the origin still on the ellipse's major axis, but shifted upward to the earth's surface. In this new coordinate system, the origin is on the ground, halfway between the two legs of the "parabola":

x^2/b^2 + (y - ae + r)^2/a^2 = 1 (eq. 2)

(where "r" is the earth's radius).

Now, rearranging (2) (and considering only the top half of the ellipse), we get the following:

y = ae - r + a*sqrt(1 - x^2/b^2) (eq. 3)

Now, near the very tip of the ellipse, in the "parabola" section, x is much less than b. That allows us to make this approximation (by the binomial theorem):

y ~= ae - r + a(1 - x^2/(2b^2)) [for x << b] (eq. 4)

Woohoo! Equation (4) is the equation for a parabola, which is what we'd expect. Now, let's compare that to the "classical" equation for a parabolic trajectory in a uniform g-field, using the same origin as above:

y = V_y^2/(2g) - (g/(2V_x^2))*x^2 (eq. 5)

(where V_x is x-component of initial velocity; V_y is y-component of initial velocity; and g = acceleration due to gravity)

Now we just match up the polynomial coefficients between (4) and (5), to get these:

V_y^2/(2g) = a(1+e) - r (eq. 6)

g / V_x^2 = a / b^2 (eq. 7)

Now we have two equations in two unknowns, a and b (the "e" can be expressed in terms of a and b; see above). So we should be able to solve this. I tried, and I got a big messy equation, but here's what I got:

The ellipse's semimajor axis (a) is:

a = (h + r)^2 / [2(h+r) - V_x^2/g] (eq. 8)

(where h is the maximum height above the ground that the ball reaches, aka V_y^2/(2g))

The ellipse's eccentrity (e) is:

e = sqrt(1 - (V_x^2/g)*a) (eq. 9)

Finally, the distance between the earth's center and the (upper) focus is:

R_f = 2*a*e (eq. 10)

Or, if you wanted to know how high above the ground the upper focus is:

H_f = 2*a*e - r (eq. 11)

Now, if you want a REAL answer, here's all you need to do:
1) Adjust V_x to take into account the earth's rotation;
2) Plug all the numbers into equations (8), (9), (10) and (11).

I leave that as an exercise for the reader. :-)

Answer 3

At infinity, after which it comes around from the other side when the curve becomes hyperbolic. Looking at conic sections is the easiest way to visualize this.

Answer 4

kepler's laws apply to movement of planetswith sun at one of the two focii of their trajectories.