I'm calculating gravitic influence as pow(distance, -2) * mass1/mass2.
The side of the planet farthest from its neighbor would be less affected by gravity, thus the planet should rotate as it orbits. How would you find the exact amount?
I like this question. Let's try this perspective.
Any time an objects changes rotational velocity, there MUST be an acting torque on it.
If you're not familiar with torque, torque is a rotational force. Torque can be measured by multiplying the force being applied onto a lever and the length of the level (so a force applied to a long lever gives more torque than the same force applied to a shorter lever).
If an object experiences no torque, there will be no difference in its rotational velocity. That means that whatever rotational speed the object had before some zero-torque event is the exact same rotational speed it will have after that event.
Let's see if there are any sources of torque during the orbit of a spherical celestial object about a large source of gravity.
We will be using one assumption that I am not proving. For solid spherical objects that give of a field that is inverse-square, the entire sphere can be modeled as a point. This means, if for example we were going to see if the Sun should impart any rotation onto the earth, we can model the sun as a single point. If this assumption isn't satisfactory, please reply and i'll try to prove it as well (be warned, it probably involves calculus).
edit: Actually, because we have so much symmetry, i think it's possible to due this entire proof without assuming a point gravitational source.
Alright, so here is the setup:
We have large source of gravity that we will be treating like a point source, and it imparts a gravitation field onto a smaller celestial sphere. Now, I am assuming this sphere has a symmetrical density about its radius.
Let's choose an arbitrary, but small volume, somewhere in our celestial sphere. This volume is shown red in the image. For my example, it's irrelevant where you choose this volume. If you choose closer to the gravity source, it will experience a stronger force, if you choose farther from the gravity source it will have a weaker force. It doesn't matter, it's irrelevant to this proof.
Now, if we calculate the force on this volume, we find that volume is some distance from the center of the celestial sphere, and therefore any force on this volume will impart a torque onto the celestial sphere.
But hold on there! We have a dotted line from the center of our gravity source through the center of our celestial object. For ANY arbitrary volume that we pick on our celestial sphere, we can find another equal volume symmetrical opposed across the dotted line. (the volume pairs are such that a line drawn between them will intersect our dotted line and create a right angle)
Because we chose a dotted line that goes from the center of our gravity source to the center of our celestial sphere, any pair of equal volumes radially symmetrical about this dotted line will have equal density and be equidistant from our gravitational source. This means an equal force due to gravity will be acting on them.
And because we also choose these two volumes to be on opposite ends, then any torques they have will cancel.
Since there force due to gravity is equal, and since they are equalidistant from the center of our celestial sphere, their torques completely cancel.
For ANY arbitrary volume that you choose on our celestial sphere, the torque it imparts onto the sphere is completely canceled by it's opposite pair. If you sum up all the pairs you can create the entire volume of the celestial sphere, and the sum of torques will still be zero. Therefore there is zero net torque on our celestial sphere.
therefore, whatever rotation velocity the celestial sphere had when it entered the gravitation field, is the same rotational velocity it will have throughout orbit, and will be the same rotation velocity it will have if it ever leaves, unless some other outside torque acts on it.
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