Formula for gravity path of two objects

In summary, the conversation discusses finding a formula to determine the actual position of two identical objects based on their acceleration and the gravitational attraction between them. The conversation explores using a periodic path and the formula for an ellipse to solve the problem, but ultimately suggests looking into orbital motion equations for a solution. The correct formula for the dependency of time and normalized distance is given as $$ \arccos \sqrt z + \sqrt {z - z^2} = pt $$.
  • #1
zrek
115
0
Please help me to find a formula to determine the actual position of two objects, depending on time.

Imagine two identical (mass, size, etc) object, no other effect on them only the gravity of each other (Newton). Let's imagine these objects are non corporal, so they will not collide.

The beginning state: speed, v=0, distance between them: 2*a;

If there would be constant acceleration, we could use the f(t)=x=v0*t+(a/2)*t^2 to determine their actual position depending on time.

But our case is different, the acceleration is changing.
F=G*(m*m)/(d*d)

Is there a simple formula now to determine the positions of the objects?

I tried to thinking this way:

Unlike by the constant acceleration, now we will get a periodic path.
The two objects are the special case of a circle/elliptic path (the objects are circulating around each other)
The formula for the ellipse:
x=a*cos(q)
y=b*sin(q)

In our case b=0.

But now I'm stucked, because I think that the q is not representing the time. (Am I right?)

Thank you for your help!
 
Physics news on Phys.org
  • #2
q tells you how far apart they are. q = 0,π,2π are max. q = π/2, 3π/2 are min (0 separation). q is a function of time, depending on physical parameters (initial condition, acceleration).
 
  • #3
You can't really solve this as a 1d problem, because the gravitational attraction diverges at zero separation.

As a 2D problem, you end up with both objects moving along elliptical paths whose foci are at the baricenter of the system. If you look up orbital motion on Wikipedia, it will give you some equations to work with.
 
  • #4
mathman said:
q tells you how far apart they are. q = 0,π,2π are max. q = π/2, 3π/2 are min (0 separation). q is a function of time, depending on physical parameters (initial condition, acceleration).

Yes, I agree that the q is some kind of an "angle" and is a function of time, but what is the exact formula of the dependency? I'm sure that all the necessary data is given. For example the actual acceleration is also depends on the actual position of the path. I have a feeling that there should be a simple solution.
 
  • #5
If ## z ## is the normalized distance from the barycenter to a mass (normalized so that ## z = 1 ## corresponds to the initial position), the equation of motion is given by $$

\arccos \sqrt z + \frac 1 2 \sqrt {z - z^2} = pt

$$ where ## p ## is a parameter that includes the masses, the initial distance, etc.
 
Last edited:
  • #6
K^2 said:
You can't really solve this as a 1d problem, because the gravitational attraction diverges at zero separation.

Isn't it help if we say that b approaches 0?
In our case if we taking "time steps" then we can easily jump over the problematic "too close" areas by saying that the two objects just "exchanged", "mirrored". I mean that the problematic forces and/or values somehow neutralizes each other, so finally if I'd like to find out the position (which is never infinite or problametic anyway) there should be a simple formula. (maybe I'm wrong)

K^2 said:
As a 2D problem, you end up with both objects moving along elliptical paths whose foci are at the baricenter of the system. If you look up orbital motion on Wikipedia, it will give you some equations to work with.

Thanks, I'll try to find there some equations that fit to this case.
 
  • #7
voko said:
If ## z ## is the normalized distance from the barycenter to a mass (normalized so that ## z = 1 ## corresponds to the initial position), the equation of motion is given by $$

\arccos \sqrt z + \frac 1 2 \sqrt {z - z^2} = pt

$$ where ## p ## is a parameter that includes the masses, the initial distance, etc.

Looks great!
So if I reverse this formula, I can get the Z as function of time (t).
I'll check it thank you!
 
  • #8
voko said:
If ## z ## is the normalized distance from the barycenter to a mass (normalized so that ## z = 1 ## corresponds to the initial position), the equation of motion is given by $$

\arccos \sqrt z + \frac 1 2 \sqrt {z - z^2} = pt

$$ where ## p ## is a parameter that includes the masses, the initial distance, etc.

Voko, please help me to understand this formula.
If the Z goes 1 to 0, then 0 should mean that the object have reached the barycenter, right?
In this case the speed should be the highest on the path.
However the pt goes from 0 to pi/2, and the time change is the smallest in the middle (at pi/4), by this the speed seems to be not the highest at z=0.
What I understand wrong?
 
  • #9
Sorry, I made a mistake in the formula. This is the correct formula: $$ \arccos \sqrt z + \sqrt {z - z^2} = pt $$
 

What is the formula for calculating the gravitational path of two objects?

The formula for calculating the gravitational path of two objects is given by Newton's Law of Universal Gravitation: F = (G * m1 * m2) / r2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

How is the formula for gravitational path derived?

The formula for gravitational path is derived from the principles of Newton's Law of Universal Gravitation, which states that every object in the universe exerts a gravitational force on every other object. The strength of this force is directly proportional to the masses of the objects and inversely proportional to the square of the distance between them.

Can the formula for gravitational path be used to calculate the trajectory of any two objects?

Yes, the formula for gravitational path can be used to calculate the trajectory of any two objects, as long as their masses and the distance between them are known. This formula is applicable to any two objects in the universe, regardless of their size or distance.

What factors can affect the accuracy of the formula for gravitational path?

There are a few factors that can affect the accuracy of the formula for gravitational path. These include the distance between the two objects, the precision of the values used for their masses, and the presence of other nearby objects that may also exert a gravitational force on the two objects in question.

In what situations is the formula for gravitational path most commonly used?

The formula for gravitational path is most commonly used in situations involving celestial bodies, such as planets, moons, and stars. It is also commonly used in astrophysics, astronomy, and space exploration to calculate the trajectories of spacecraft and other objects in space.

Similar threads

Replies
9
Views
2K
  • Other Physics Topics
Replies
6
Views
857
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
276
Replies
10
Views
924
  • Other Physics Topics
Replies
4
Views
1K
  • Other Physics Topics
Replies
9
Views
2K
Replies
19
Views
1K
Replies
1
Views
559
  • Other Physics Topics
Replies
4
Views
1K
Back
Top