I don't think that's quite right. Your equation supposes that the mass is all at a singular point and that you are suspended/supported on a massless sphere 1.5 earth radii away. To figure it out precisely, you may need to include rotation speed as well (depending upon whether it is fast enough to be significant).
You’re right about the angular momentum. THe planet’s round though. So I’d expect the order of magnitude for that kind of a correction to be smaller by less than 2. Earth’s is ~0.3% at the equator.
Actually, spunkets' formula is completely correct. It doesn't matter whether the planet has a 10 mile radius or a radius of 1.5 Earths (roughly 6,000 miles), if you are at a distance of 1.5 Earths from the center of mass, the gravitational pull is the same. As spunkets pointed out, the inverse square law applies.
It also doesn't matter that all planets are denser toward the center, than near the surface. Gravitational anomalies (masses at the same depth with different densities) have a slight effect, but they would not be noticable to you.
Rotation would also be insignificant, unless the planet has a very fast rotational velocity. On the Earth's Equator, you are travelling around the Earth at about 1,000 miles per hour, but you can stand at the Equator, or at one of the Poles, and not notice the difference in weight (you are slightly heavier at the Poles).
This is simple Newtonian Statics and Dynamics, which is normally covered in the first quarter of college physics. Those who take physics in high school are likely to learn this before they get to college.