This is what I am trying to understand.
Radial velocity is the technique used for detecting most extrasolar planets (transit detection is rare). I can understand how more a more massive planet will cause a larger shift in the stars radial velocity than a less massive planet in an identical orbit.
It is also apparent that a shorter period orbit would be detected more easily than a longer period orbit because the required observation time is shorter, and the planet exerts more pull on the star when they are close together.
This is where it gets messy: Two equally massive planets, one in a circular orbit, the other in an elliptical orbit with equal perihelia. The peak excursion in the stars radial velocity should be the same for both, but the period of the eccentric orbit will be longer, reducing it's detectability.
In addition, we can only detect the component of radial velocity along the line of sight to the star. As the semi-major axis approaches perpendicularity to the line of sight, detectability decreases.
What am I missing here?
Remember when a planet is orbiting a star; it also has a gravitational influence on that star as well. So it’s not just a planet orbiting a star, it's both orbiting a common center of mass.
So as the planet becomes more massive, the further the center of mass moves from the center of the star (causing the star to wobble).
Also the period of an orbit is directly proportional to its distance from the star, closer is shorter.
So the massive close-in higher elliptical planets are far easier to detect.
If a planet has a perfectly circular orbit, the center of mass of the system will rotate around the center of the star in a uniform fashion and usually does not cause the star to wobble enough to be detectable by our present method of stellar displacement.
So in conclusion, a massive planet with a close in higher elliptical orbit will cause an appreciable fast enough wobble that is easier to detect.
Physicist, longshadow, MikeD; All; Anything you want to add?