The eccentricity is 1.0.
The eccentricity $e$ of an orbit can be found from the radius of apoapse and periapse as:
and the semimajor axis $a$ can as well, from:
If you throw an object horizontally (velocity perpendicular to position vector) you will end up in a closed orbit if you throw at slower than escape speed, an open parabolic orbit if you throw it exactly at escape speed, or an open hyperbolic if you throw it greater than escape speed.
There is a speed which will result in a perfect circular orbit, with $e=0$. We might as well call this "circular orbit speed". In the eccentricity equation, if $r_a=r_p$ as it does in a circular orbit, we see that the numerator is zero, while the denominator is nonzero, so the whole fraction is also zero.
If you throw slower than circular velocity, the object will fall closer to the center before coming back up. The lower the object gets, the lower periapse is. In the eccentricity equation, as $r_p$ decreases, the numerator grows while the denominator shrinks, so the whole fraction increases. As we go slower, we increase eccentricity.
The limiting case of this is if you throw it at zero speed, IE you drop it.
For an object dropped in a gravity field around a true point mass, you will end up with the apoapse being the radius at which you dropped the object, and the periapse at zero. This is a very weird orbit, because the object will take a finite time to reach the center, but will reach infinite speed just as it passes the center where it will make a 180° turn and coast back up, until it reaches its original drop height at zero speed and starts another cycle. You can use Kepler's third law to figure the time of this orbit, since it still has a well-defined $a$.
A spherically-symmetric mass with a definite surface (density of zero outside a certain radius) has an identical gravity field to that of a centered point mass everywhere outside its surface. Therefore an object dropped above the surface on a more-realistic planet would follow an orbit identical to that dropped at the same radius above a point mass, until it hit the surface. If it were to pass through the surface (say you drilled a hole) the gravity field below the surface is not the same as that of a point mass.
Whenever I am at a baseball game and see a pop-fly, it always amuses me to think that the path the ball is following is not truly a parabola, just the end of a very stretched out ellipse, which if continued, would form the same shape near the center of the Earth.
In this case, $r_p=0$. The eccentricity fraction has its numerator equal to $r_a-0=r_a$, and denominator equal to $r_a+0=r_a$ as well. The eccentricity is exactly 1.0 .
"But Kwan!" I hear you shout. "If $e=1$, doesn't that make it a parabolic orbit?" In this case, no. A parabolic orbit has $e=1$ and $a=\infty$, while the drop orbit has $e=1$ but a decidedly non-infinite $a=r_a/2$.
This case is the limit of an ellipse becoming thinner and thinner as the foci move apart. In that limit, one focus is at the center, one at the drop point, and the ellipse has zero width but finite length.
In this image, the circle that isn't moving represents a sphere with radius 1, and the ellipse that is moving represents an orbit with a constant apoapse radius of 2.0 but a varying eccentricity.
Note that this is not what would happen if you drilled a hole through the earth and dropped an object. This only applies to a true point mass, and doesn't take into account relativity (a true point mass would be a singularity, and the object would pass the event horizon on the way down and never come back up).