Falling object far from Earths Center of Mass Problem

In summary, the conversation discusses the speaker's struggle with understanding and finding a function of time for the n-body problem in physics. They mention their understanding of gravity and acceleration and how it changes over distance, as well as the need for a function of time in order to accurately calculate the position and velocity of an object. The speaker also mentions the complexity of the problem and the lack of literature available on the topic. They suggest using numerical methods and differential equations to solve the problem.
  • #1
I have been thinking about how to come up with a function of time for the n-body problem. Later I found that there still is no function of time known for the n-body problem, but I am still not sure where the brick wall for everyone is because I am quite sure my understanding has hit from very early on. I pretty much started from the ground up, where gravity is constant, and barely got through the basics. There are no explanations for what I am looking for to be found in any of my physics books, calculus book, or on the internet. I feel it is very basic yet I can't quite mathematically get anywhere, even though all of the concepts I understand clearly.

So from the start to where I am stuck:

-For a constant acceleration of gravity, such as near the Earth's surface where the acceleration of gravity = "g" [OPEN attachment A]
-Then integrate it to make the height as a function of time. [OPEN attachment B]

It is Easy so far, but when you make the height considerably far from Earth and consider it over a long fall toward earth, the Gravitational Force on the object is constantly changing and the function of time would have to respond to a function of constantly changing acceleration:
where: F = ma = GmM(1/r[1/r^2])

divide by m : a = F/m = (GM)/r^2 !

Now if we sum up the entire acceleration experienced for the object from the "initial r" to the "final r", we would simply integrate that function to get: [OPEN attachment C]

So the final velocity at any "final r" is:
V("final r") = initial V + GM[(1/ "final r") - (1/ "initial r")

This is as far as my understanding goes and now I would like to understand how to make this a function of time. Surely NASA can do something like this. They obviously know where all of the planets will be around the sun at any point in time and they obviously must have an understanding on how to send probes all over the solar system, orbiting various planets at various times and sling-shotting around this and that. Yet I can not find any literature on any of this.

ANY HELP WOULD BE GREAT. Once I got started on this not knowing how to do it has been driving me insane!


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  • #2
Now if we sum up the entire acceleration experienced for the object from the "initial r" to the "final r", we would simply integrate that function to get:
V("final r") = initial V + GM[(1/ "final r") - (1/ "initial r")
No, because the velocity would be the integral with respect to time, and you integrated the right hand side with respect to r.

What you want to do is look at the conservation of energy:

½mv2 + GMm/r = the same for initial and final states
  • #3
To give you an idea of how complext this is, the math for 2 point objects starting at rest is shown in post #8 with minor correction (3 intermediate steps) in post #15 of this thread:


The math for fixed size objects by arildno in post 2 and 4 in this thread but the latex is messed up in post #2 (cleaned up in a later thread see below):


arildno's math with the latex cleaned up is in post #3 of this thread:


The issue for a n body system is probably related to stating the paths (position versus time) of the objects in the form of an equation. This wouldn't prevent using numerical methods based on differential equations to solve the problem.
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  • #4
You want to describe for example when a asteroid its falling to the Earth but from the outer space?. In this case, the gravity its not g, like you say, but its function of the distance from the objetc to the center of the earth.

with R the radius of the earth, M the mass of the Earth and G the universal constant and m the mass of the objetct

if you want to deduce the differential equation we must put the 2nd law of Newton

[itex]\sum F=m_{2}\ddot{y}=\dfrac{GM_{1}m_{2}}{(R+y(t))^{2}}[/itex]

and if you want to put drag forces or rotational forces, put it and describe them. finally you must solve de second order ode
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  • #5

I can understand your frustration with trying to find a function of time for the n-body problem. This problem has been a topic of research and debate for many years and, as you mentioned, there is still no known solution. The n-body problem refers to the mathematical problem of predicting the motion of a group of celestial bodies, such as planets, stars, and satellites, under the influence of their mutual gravitational attraction.

One of the main challenges in solving this problem is that the gravitational force between two objects is not constant, but rather depends on their changing distances and positions. This makes it difficult to come up with a single function of time that can accurately predict the motion of these objects. Additionally, as the number of objects increases, the complexity of the problem also increases, making it even more challenging to find a solution.

While NASA and other space agencies are able to accurately predict the positions and movements of celestial bodies, they do so using advanced computer simulations and mathematical models, rather than a single function of time. These models take into account various factors such as the gravitational force, the mass and velocity of the objects, and other external forces.

I encourage you to continue your research and exploration of this problem, as it is a fascinating and important area of study in physics. However, it is also important to keep in mind that even the most advanced scientists and researchers are still working on finding a solution to the n-body problem. Keep pushing yourself to learn and understand more, but also be patient and know that this is a complex and ongoing challenge in the scientific community.

1. How does the distance from Earth's center of mass affect the falling object?

The distance from Earth's center of mass has a direct impact on the gravitational force acting on the falling object. The farther the object is from the center of mass, the weaker the gravitational force will be. This means that the object will fall slower and take longer to reach the ground compared to an object closer to the center of mass.

2. What is the formula for calculating the gravitational force between two objects?

The formula for calculating the gravitational force between two objects is F = G * (m1 * m2) / d^2, where F is the force of gravity, G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and d is the distance between their centers of mass.

3. How does the mass of the falling object affect its acceleration?

The mass of the falling object does not affect its acceleration due to gravity. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Since the force of gravity remains constant regardless of the object's mass, the acceleration due to gravity also remains constant.

4. What other factors can affect the falling object far from Earth's center of mass?

The main factor that can affect the falling object far from Earth's center of mass is air resistance. As the object falls, it will encounter air molecules that will push against it and slow it down. This can affect the object's speed and time it takes to reach the ground.

5. Can the falling object ever escape the Earth's gravitational pull?

Yes, if the falling object has enough initial velocity, it can overcome the Earth's gravitational pull and escape into outer space. This is known as escape velocity and is dependent on the mass and radius of the planet. However, for most objects, the Earth's gravitational pull is strong enough to keep them from escaping.

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