


01162013, 08:48 AM

#1

Diamond Member
Join Date: Jan 2013
Location: Brooklyn, NY
Posts: 9,810

Calculating a planet's rotation
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?



01162013, 12:57 PM

#2

Lifer
Join Date: Oct 2002
Posts: 10,569

First point... great drawing.
Next point, I didn't think there was a requirement that all planets must rotate in steady state when orbiting another mass. Take the moon and earth for example. So I guess if you could clarify the "thus the planet should rotate" part. I'm rusty on physics so I could be completely off base.
__________________
post count = post count + 0.999.....
(\__/)
(='.'=)This is Bunny. Copy and paste bunny into your
(")_(")signature to help him gain world domination.



01162013, 01:19 PM

#3

Diamond Member
Join Date: Nov 2000
Posts: 4,821

Quote:
Originally Posted by Rakehellion
The side of the planet farthest from its neighbor would be less affected by gravity, thus the planet should experience tidal bulging along the axis of gravitational pull (i.e. the axis that constantly changes orientation to stay pointed in the direction of the other gravitating body).

fixed. the fact that the gravitational force exerted by a body on another body's far side is less than the gravitational force exerted on that same body's near side does not imply rotation. if anything, such a gravitational interaction would tend to slow the rotation of a body, provided it is close enough to its gravitating neighbor. taking it one step further, if the bodies are similar in mass, then the rotation of both bodies will tend to slow over time (again, provided they're close enough in proximity), eventually resulting in a tidal lock.
Quote:
Originally Posted by TuxDave
I didn't think there was a requirement that all planets must rotate in steady state when orbiting another mass. Take the moon and earth for example.

but if you think about it, the moon does rotate...its rotational period just happens to be the same as its orbital period. because its rotation period matches its orbital period, one side of the moon always faces earth, and the other side always faces away from earth.
__________________
1) i7 3770K, Hyper 212+, Gigabyte G1.Sniper 3, 2 X HD 7970, SeaSonic X750, Win7 x64
2) i7 3770K, Hyper 212+, Gigabyte G1.Sniper 3, 2 X HD 7970, SeaSonic X750, Win7 x64
3) i7 3770K, Hyper 212 EVO, Gigabyte Z77XUD4H, HD 7970, SeaSonic X650, Win7 x64
4) X6 1055T, Hyper 212+, MSI 890GXMG65, 2 X HD 6670, Antec Neo Eco 620, Win7 x64
My Heat (5600)  BOINCstats



01162013, 01:33 PM

#4

Lifer
Join Date: Oct 2002
Posts: 10,569

Quote:
Originally Posted by Sunny129
but if you think about it, the moon does rotate...its rotational period just happens to be the same as its orbital period. because its rotation period matches its orbital period, one side of the moon always faces earth, and the other side always faces away from earth.

Woops, that's right.
__________________
post count = post count + 0.999.....
(\__/)
(='.'=)This is Bunny. Copy and paste bunny into your
(")_(")signature to help him gain world domination.



01162013, 05:37 PM

#5

Golden Member
Join Date: Feb 2008
Posts: 1,563

Quote:
Originally Posted by Sunny129
fixed. the fact that the gravitational force exerted by a body on another body's far side is less than the gravitational force exerted on that same body's near side does not imply rotation. if anything, such a gravitational interaction would tend to slow the rotation of a body, provided it is close enough to its gravitating neighbor. taking it one step further, if the bodies are similar in mass, then the rotation of both bodies will tend to slow over time (again, provided they're close enough in proximity), eventually resulting in a tidal lock.

That kind of damping is due to deformation of the bodies, so the normal rigid physics model OP is using would not model it. Just pointing this out to prevent confusion.



01162013, 07:58 PM

#6

Administrator Elite Member Goat Whisperer
Join Date: Mar 2001
Location: Western NY
Posts: 46,761

Just to be technical, (given the forum), inertia isn't a vector. You're confusing it with momentum. Inertia is equivalent to mass.
Wikipedia's entry for tidal locking includes a formula to estimate how much time would elapse before two bodies became tidally locked.



01162013, 08:17 PM

#7

Diamond Member
Join Date: Jan 2013
Location: Brooklyn, NY
Posts: 9,810

Quote:
Originally Posted by Sunny129
but if you think about it, the moon does rotate...its rotational period just happens to be the same as its orbital period. because its rotation period matches its orbital period, one side of the moon always faces earth, and the other side always faces away from earth.

And if the body escapes orbit, it should continue rotating. Or if the body wasn't rotating before it met its neighbor, it will begin.
I'm needing to calculate these values on the computer, so my numbers are appropriately naive and parsimonious.
Is there a lazy way to plug Θ, mass1, mass2, diameter1, diameter2, and distance into a function to approximate the angle of rotation?



01182013, 02:56 PM

#8

Golden Member
Join Date: Feb 2008
Posts: 1,563

Quote:
Originally Posted by Rakehellion
I'm needing to calculate these values on the computer, so my numbers are appropriately naive and parsimonious.

???
Gravity is a force. Inertia is not a force. "Gravity + inertia" is nonsense, and any angle dependent on it is likewise nonsense.
Quote:
Is there a lazy way to plug Θ, mass1, mass2, diameter1, diameter2, and distance into a function to approximate the angle of rotation?

There is no way, whether easy or hard. If you are working with a rigid physics model, the bodies do not change the rotation of each other and will just keep rotating at a constant rate. And if it's supposed to be a nonrigid model, then you haven't given any parameters at all describing the nonrigid part.



01192013, 10:02 PM

#9

Diamond Member
Join Date: Jan 2013
Location: Brooklyn, NY
Posts: 9,810

Quote:
Gravity is a force. Inertia is not a force. "Gravity + inertia" is nonsense, and any angle dependent on it is likewise nonsense.

Kinetic energy of the planet's current motion. I'm not really trying to calculate force, I'm trying to calculate movement in response to force.
Quote:
If you are working with a rigid physics model, the bodies do not change the rotation of each other and will just keep rotating at a constant rate.

How is that possible when the force exerted on them is not constant?
Last edited by Rakehellion; 01192013 at 10:06 PM.



01202013, 09:37 AM

#10

Diamond Member
Join Date: Aug 2005
Posts: 3,480

Quote:
Originally Posted by DrPizza
Just to be technical, (given the forum), inertia isn't a vector. You're confusing it with momentum. Inertia is equivalent to mass.

That's not a simple technicality!
OP, you need to read a physics book. Your illustration is so far from making any sense that you might as well plot apples vs. oranges.
Last edited by KillerCharlie; 01202013 at 11:52 AM.



01202013, 10:49 AM

#11

Senior Member
Join Date: Jan 2010
Posts: 276

Uranus has an axial tilt of 97 degrees. The planet rotates on its 'side' so the north pole will experience 42 years of sunlight, followed by 42 years of darkness.
Venus' axial tilt is 177 degrees. It rotates in the opposite direction compared to most of the other planets.
How can you account for those two planets in your model?



01202013, 01:20 PM

#12

Junior Member
Join Date: May 2012
Posts: 5

Afaik a planets rotation exists because of conservation of angular momentum. If I remember correctly applied force will either add to this or subtract from it. As somebody pointed out earlier the effect you are trying to describe will subtract from the angular momentum, and thus slowing the rotation.
Trying to clarify what I think you have problems understand I can say this, linear motion (which is usually easy to understand) has analogy to angular motion like this:
Mass > Interia
Force > Torque
Mass and inertia are not vectors, force and torque is.



01202013, 02:28 PM

#13

Diamond Member
Join Date: Jan 2013
Location: Brooklyn, NY
Posts: 9,810

Quote:
Originally Posted by KillerCharlie
That's not a simple technicality!
OP, you need to read a physics book. Your illustration is so far from making any sense that you might as well plot apples vs. oranges.

I forgot to mention that I'm doing this in real time on a computer, so what I need is a very close approximation. I can't model every particle and I probably can't do nonrigid bodies.
Despite my mixing of terminology, several people seemed to understand what I meant just fine. I'm posting here apparently because I'm not familiar with the physics involved. I need help, not putdowns.
Quote:
Originally Posted by mpo
Uranus has an axial tilt of 97 degrees. The planet rotates on its 'side' so the north pole will experience 42 years of sunlight, followed by 42 years of darkness.
Venus' axial tilt is 177 degrees. It rotates in the opposite direction compared to most of the other planets.
How can you account for those two planets in your model?

I don't. The idea is that I'll create my own solar system wherein orbit, rotation, and axial tilt will render themselves organically.
Quote:
Just to be technical, (given the forum), inertia isn't a vector. You're confusing it with momentum. Inertia is equivalent to mass.

I meant velocity sustained by inertia. Can't get too verbose in iPhone sketches.
Planet A moves in a path past Planet B
The far side of Planet A will be less affected by gravity, thus continue moving along the path
The near side of the planet will be more compelled to follow gravity
The near and far sides of the planet are moving in different directions, thus causing it to rotate
Right? I need to know by how much. I don't know anything about the calculus involved and it would help immensely if someone could point me in the right direction.



01202013, 04:00 PM

#14

Diamond Member
Join Date: Aug 2005
Posts: 3,480

I am not trying to put you down, I am trying to point you in the right direction. I work as an aerospace engineer, have taken multiple orbital mechanics courses, and tutored college physics. I can tell that you are struggling with some very basic concepts, like the difference between force and mass. You simply cannot try to jump to an understanding of more advanced concepts without understanding the basics.
First of all, if you make a sketch with vectors on it, they have to be indicating the same thing. Plotting inertia and a force on the same plot is not useful. Inertia is not even a vector, it's essentially a measurement of mass like Dr. Pizza said.
You're also trying to get a rotation out of a point mass  you're drawing forces that are acting at that point and there's no torque, thus no source of angular momentum. The only force acting in your sketch and modelling simplification is gravity. That's it. From that force, and a few starting values, you can easily compute the orbital speed of one body around another, but not its rotational velocity at all.
In fact, It is 100% impossible to predict the rotational speed of a planetary body short of directly measuring it.
In your model you're assuming constant density, uniform planets. A spherical, uniform planet behaves exactly like an infinitely small point mass (when you integrate the gravitational forces over its volume). Sure, the far side of the planet has less gravitational force, but the close side has more, balancing it out. No rotation is created.
If you take a massive leap in the fidelity of you're modelling, it is possible to model the nonspherical, nonconstant density planet. Because the planet isn't uniform, a small torque could be imparted on a planet, changing its angular velocity  but the angular momentum of the system is constant without outside force. It is possible to compute this torque with detailed data, but it won't tell you anything about the current rotational speed of the planet. Tidal forces and such can damp the rotation of a body, but the total angular momentum of the system is still conserved.
The rotational speed of a planet about its axis is mostly a function of how the planet was formed (how the space dust came together) and discrete events like asteroid impacts. There's nothing that says the planets even have to spin in the same direction  some do not (retrograde motion). The spin of a planet is mostly unrelated to its orbit.
And the "computer program" you're using  I can tell you as a practicing engineer that unless you understand exactly what a computer program is doing, it will do you more harm than good 100% of the time. You need to understand the physics and math behind the problem, starting with the basics. There is no shortcut.
Last edited by KillerCharlie; 01202013 at 05:13 PM.



01212013, 12:24 PM

#15

Diamond Member
Join Date: Jan 2013
Location: Brooklyn, NY
Posts: 9,810

Quote:
In your model you're assuming constant density, uniform planets. A spherical, uniform planet behaves exactly like an infinitely small point mass (when you integrate the gravitational forces over its volume). Sure, the far side of the planet has less gravitational force, but the close side has more, balancing it out.

Why would it behave as a point? Wouldn't uneven forces make things less balanced, not more?
A cannon ball is fired from a cannon, hits the ground, and begins to roll in the direction of travel. What's wrong with this analogy?
Quote:
In fact, It is 100% impossible to predict the rotational speed of a planetary body short of directly measuring it.

That sounds like an exaggeration.
Last edited by Rakehellion; 01212013 at 12:30 PM.



01212013, 01:42 PM

#16

Golden Member
Join Date: Feb 2008
Posts: 1,563

Quote:
Originally Posted by Rakehellion
Why would it behave as a point? Wouldn't uneven forces make things less balanced, not more?
A cannon ball is fired from a cannon, hits the ground, and begins to roll in the direction of travel. What's wrong with this analogy?

The friction between the cannon ball and the ground is only on one side of the ball, thus there's nonzero torque. In the case of rigid planets and gravity, if you consider the line that goes through the centers of the planets, you can mirror every point of a planet about that line to find another point whose torque from gravity cancels out the torque from the first point. Thus, the rotation cannot change.



01222013, 08:48 PM

#17

Diamond Member
Join Date: Aug 2005
Posts: 3,480

Quote:
Originally Posted by Rakehellion
That sounds like an exaggeration.

Quote:
Originally Posted by Wikipedia
"As this interstellar dust is inhomogeneous, any asymmetry during gravitational accretion results in the angular momentum of the eventual planet. The current rotation period of the Earth is the result of this initial rotation and other factors"

It is not possible to calculate the rotation of a planet without measuring it directly. What other proof do you need?
I can pull out my textbooks and show you the derivation if you really want to see it. However, that depends on understanding calculus and physics, so it'll just be nonsense to you.



01232013, 02:32 PM

#18

Lifer
Join Date: Jul 2001
Location: London, UK
Posts: 27,710

Quote:
Originally Posted by KillerCharlie
It is not possible to calculate the rotation of a planet without measuring it directly. What other proof do you need?
I can pull out my textbooks and show you the derivation if you really want to see it. However, that depends on understanding calculus and physics, so it'll just be nonsense to you.

The fact is the rotational period of a planet will change throughout time too. It can't just be predicted.
Some planets spin quickly, some are tidally locked, some spin "backwards", some spin on their side.



01232013, 04:03 PM

#19

Senior Member
Join Date: Jan 2012
Posts: 545

Quote:
Originally Posted by Rakehellion
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 zerotorque 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 inversesquare, 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.
Last edited by serpretetsky; 01232013 at 04:18 PM.



01232013, 04:40 PM

#20

Golden Member
Join Date: Feb 2008
Posts: 1,563

I kinda said that already, but with one sentence.



01232013, 06:36 PM

#21

Lifer
Join Date: Jul 2001
Location: London, UK
Posts: 27,710

Quote:
Originally Posted by serpretetsky
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.

That only works for perfect spheres that are perfectly rigid... which planets are not. Tidal forces will transfer energy until the planet is in tidal lock.



01232013, 06:38 PM

#22

Diamond Member
Join Date: Nov 2000
Posts: 4,821

Quote:
Originally Posted by Pia
I kinda said that already, but with one sentence.

yes, but nothing drives the point home quite like an illustration, and as someone who is fairly novice in the physics department, i think this may be exactly what the OP needs to see in order to understand why gravitation does not induce rotation of the body being acted upon. we'll see how he responds...
__________________
1) i7 3770K, Hyper 212+, Gigabyte G1.Sniper 3, 2 X HD 7970, SeaSonic X750, Win7 x64
2) i7 3770K, Hyper 212+, Gigabyte G1.Sniper 3, 2 X HD 7970, SeaSonic X750, Win7 x64
3) i7 3770K, Hyper 212 EVO, Gigabyte Z77XUD4H, HD 7970, SeaSonic X650, Win7 x64
4) X6 1055T, Hyper 212+, MSI 890GXMG65, 2 X HD 6670, Antec Neo Eco 620, Win7 x64
My Heat (5600)  BOINCstats
Last edited by Sunny129; 01232013 at 06:45 PM.



01242013, 03:22 AM

#23

Senior Member
Join Date: Jan 2012
Posts: 545

Quote:
Originally Posted by silverpig
That only works for perfect spheres that are perfectly rigid... which planets are not. Tidal forces will transfer energy until the planet is in tidal lock.

Excuse me, i thought we were simply talking about rigid physics since the OP already admitted can't do non rigid physics.



01252013, 11:52 AM

#24

Lifer
Join Date: Jul 2001
Location: London, UK
Posts: 27,710

Quote:
Originally Posted by serpretetsky
Excuse me, i thought we were simply talking about rigid physics since the OP already admitted can't do non rigid physics.

Right you are then.



01262013, 03:51 PM

#25

Diamond Member
Join Date: Jan 2013
Location: Brooklyn, NY
Posts: 9,810

Quote:
Originally Posted by serpretetsky
Excuse me, i thought we were simply talking about rigid physics since the OP already admitted can't do non rigid physics.

I said I might be able to, but processor power is very limited. I want to simplify the model as much as possible.
I see that my original notion was way off and I'm looking into tidal locking.



Posting Rules

You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is Off



All times are GMT 5. The time now is 11:15 AM.
