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.
So, you're saying that if you were to dig a hole to within 10' of the center of the earth (allowing that you'd have to reinforce the walls), that the subjective force would be greater than at the surface? Yes, the total gravity exerting itself on you would be greater, but a portion (in this case, a large one) would be countered by opposing gravitational force. One isn't attracted by a center of gravity, but by each individual bit of mass in the universe, including the planet portions lateral to you and (in the example) above you. Gravimetric studies on Earth will sometimes make use of the effect by being conducted adjacent to large dense mountains.
As spunkets pointed out, the inverse square law applies.
While my skipping past my original off-the-cuff-statement referencing 5Gs may suggest otherwise, I agree. I was merely saying that it seemed incomplete.
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.
That does make it different, more like the idealized example I gave.
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).
The rotation effect may indeed be important. What that rotation is at the stated distance from its star is largely dependent upon the age of the system - which I don't have the information to estimate. Large bodies close-in to other large bodies have a tendency to rotate surprisingly fast - potentially much faster than earth. On the other hand, in an old system, the planet could have transferred its rotational energy to its orbital energy in the same way that our Moon appears to have, making it tidally locked. Most probably it is somewhere in between. Where, I couldn't say.
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.
No need to get snooty, snippy, and snotty. My logic problem was much more fundemental than that: It's been a while since I formally studied physics and so I worked from a mental model of an idealized mass. The Earth and nearly all other significant bodies are spheroids or discs, so the issue I had in mind was already primarily taken into account.