Distance function in gravitational force

In summary: E_k=\lim_{x\rightarrow \propto }{G\frac{Mm}{x}}+C=0So when the distance of the two particles equal to infinity, kinetic energy is zero.So C=0.Hi,In summary, the conversation discusses finding the distance function dependent on time in the context of gravitational force law. The conversation covers integrating the equation of motion, discussing acceleration in the radial and tangential directions, and the role of a constant of integration in the resulting equation. It is ultimately concluded that it is impossible to express the distance function in terms of time using elementary functions.
  • #1
coki2000
91
0
Hi,
In gravitational force law [tex]F=-G\frac{M_{1}M_{2}}{x^{2}}[/tex]if i want to find the distance function which depends on time then

[tex]\frac{d^{2}x}{dt^{2}}=-G\frac{M}{x^2}[/tex]

[tex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-G\frac{M}{x^2}[/tex]

[tex]vdv=-G\frac{M}{x^2}dx[/tex]

integrate the both sides

[tex]\frac{v^2}{2}=G\frac{M}{x}\Rightarrow \frac{dx}{dt}=\sqrt{2G\frac{M}{x}}\Rightarrow dt=\sqrt{\frac{x}{2Gm}}dx[/tex]

if we solve that differential equation then [tex]t=\frac{2}{3}\sqrt{\frac{1}{2GM}}x^{3/2}[/tex]

this time distance function [tex]x(t)=\sqrt[3]{\frac{9}{2}GM}.t^{2/3}[/tex] Is that function right else what is the right formula? Please help me
 
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  • #2
Good morning,

Your TeX formulas are not being displayed in image form.
You have to put [tex][tex][/tex][/tex] tags around the TeX formulas so that they are displayed correctly. Like this:
[tex]\textbf{[tex]}[/tex]F=-G\frac{M_{1}M_{2}}{x^{2}}[/tex].
I don't know how to answer your question, but I will follow this post, because I'm also very interested in knowing.
 
  • #3
I have never studied this subject in depth (I'm also very curious to know the solution for this), but I think that, when you integrate both sides, you forgot to add the constants of integration:
[tex]\frac{v^2}{2}+C_1=G\frac{M}{x}+C_2\Rightarrow \frac{dx}{dt}=\sqrt{2G\frac{M}{x}+C}\Rightarrow dt=\sqrt{\frac{x}{2Gm+Cx}}dx[/tex]
where C1 and C2 are the constants of integration of both sides, and, for simplicity, I use C = 2C2 - 2C1.
Let's wait for someone else to give an opinion.
Thank you in advance.
 
  • #4
coki2000 said:
Hi,
In gravitational force law [tex]F=-G\frac{M_{1}M_{2}}{x^{2}}[/tex]
[...]
I want to find the distance function which depends on time [...]

What you are looking for is a solution to the equation of motion, for gravitational force, in polar coordinates.
I will present a discussion of how to arrive at the equation of motion in polar coordinates, I don't know an analytical solution to it.

Acceleration in radial direction
For any object confined to circumnavigating motion, at each point in time the magnitude of the required centripetal force is [tex]m \dot\theta^2 r[/tex]
If that centripetal force is not provided the object will recede from the center of attraction with an acceleration of [tex]\dot\theta^2 r[/tex]

Gravitational attraction will tend to reduce the radial distance, so it gets a minus sign in the equation of motion for the acceleration in radial direction.

[tex] \ddot r = -G M r^{-2} + \dot\theta^2 r [/tex]Acceleration in tangential direction
Since it's orbital motion angular momentum is conserved. Angular momentum being constant means that the time derivative of angular momentum must be zero. Angular momentum is proportional to [tex] \dot\theta r^2 [/tex]

[tex]\dot\theta r^2 = constant \qquad \Rightarrow \qquad \frac{d(\dot\theta r^2)}{dt} = 0 [/tex]

Differentiating, using the product rule:

[tex] r^2 \ddot\theta + \dot\theta \frac{d(r^2)}{dt} = 0 [/tex]

Using the chain rule an expression with a factor r² is converted to an expression with a factor r.

[tex]r^2 \ddot\theta + 2 r \dot\theta \dot r = 0 [/tex]

Rearranging and dividing left and right by r², we obtain the equation for tangential acceleration:

[tex]\ddot\theta = - \frac{2 \dot\theta \dot r}{r} [/tex]

Notated as a system of equations:

[tex] \ddot r = -G M r^{-2} + \dot\theta^2 r [/tex]

[tex]\ddot\theta = - \frac{2 \dot\theta \dot r}{r} [/tex]

To find, as a function of time, the distance of the orbiting object to the primary that system of equations must be solved.

Cleonis
 
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  • #5
coki2000 said:
Hi,
In gravitational force law [tex]F=-G\frac{M_{1}M_{2}}{x^{2}}[/tex]
[...]
I want to find the distance function which depends on time [...]

Oops.

I assumed that the question was for orbiting motion, but it doesn't say so anywhere.
I should have assumed that the question is about falling straight down, in a situation where the attracting force is a function of distance.

Cleonis
 
  • #6
coki2000 said:
integrate the both sides

[tex]\frac{v^2}{2}=G\frac{M}{x}[/tex]

Where is your constant of integration? Once you add it in, you'll find that the resulting equation is identical to the conservation of energy. You'll also find that it's impossible to express x in terms of t using elementary functions.
 
  • #7
In my opinion, I don't forget the constant;

[tex]vdv=-G\frac{M}{x^2}dx[/tex] integrating both sides

[tex]\frac{v^2}{2}=G\frac{M}{x}+C[/tex]

But multiply the both sides by m

[tex]\frac{1}{2}mv^2=G\frac{Mm}{x}+C[/tex] this is kinetic energy formula then,

[tex]\frac{1}{2}mv^2=E_k=\lim_{x\rightarrow \propto }{G\frac{Mm}{x}}+C=0[/tex]

So when the distance of the two particles equal to infinity, kinetic energy is zero.So C=0.
 
  • #8
Usually you assume that the two particles start off at a certain distance r0 (x=r0) with speed v=0. How can you assume the two particles start at infinity?

You left off the constant of integration again in your second integral:

coki2000 said:
[tex]\frac{v^2}{2}=G\frac{M}{x}\Rightarrow \frac{dx}{dt}=\sqrt{2G\frac{M}{x}}\Rightarrow dt=\sqrt{\frac{x}{2Gm}}dx[/tex]

if we solve that differential equation then [tex]t=\frac{2}{3}\sqrt{\frac{1}{2GM}}x^{3/2}[/tex]

If you assume that the x=infinity when t=0 like you did for the first integration, you get 0=infinity+C and C=-infinity. Not a meaningful result.
 
  • #9
coki2000 said:
[tex]\frac{1}{2}mv^2=E_k=\lim_{x\rightarrow \propto }{G\frac{Mm}{x}}+C=0[/tex]

So when the distance of the two particles equal to infinity, kinetic energy is zero.So C=0.
You don't know that. You are arriving at C=0 by assuming that [tex]v_{\infty} = 0[/tex], and that is not a valid assumption.
 
  • #10
ideasrule said:
Usually you assume that the two particles start off at a certain distance r0 (x=r0) with speed v=0. How can you assume the two particles start at infinity?

You left off the constant of integration again in your second integral:

If you assume that the x=infinity when t=0 like you did for the first integration, you get 0=infinity+C and C=-infinity. Not a meaningful result.
I assume two particles which effect only gravitational force.Then they have kinetic energy by the force.If the force is absent(or infinitly small) then kinetic energy is not existing.If [tex]{C}\neq0[/tex] then kinetic energy[tex]{E_k}\neq{0}[/tex] when distance is infinity.If [tex]{C}\neq0[/tex] then the formula is wrong.So it may be zero.
 
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  • #11
coki2000 said:
If the force is absent(or infinitly small) then kinetic energy is not existing.
A particle moving in empty space (no force whatsoever) with non-zero velocity has non-zero kinetic energy. The presence of a force is not a necessary condition for something to have kinetic energy.
 
  • #12
Okey.I understand my mistake.Actually the main reason of my exploration is that I want to know why the distance and force are inversely proportional and why second power of distance.Because in many formula (Coulomb's law and others) the distance and force are inversely proportional and i think if i know it, i understand many problems in my mind.Is the main reason phenomenological or from a differential equation.Please explain to me.
Thanks.
 

What is a distance function in gravitational force?

A distance function in gravitational force is a mathematical representation of the relationship between the distance between two objects and the strength of the gravitational force between them. It is used to calculate the magnitude of the force between objects based on their distance apart.

How is the distance function in gravitational force calculated?

The distance function in gravitational force is calculated using the formula F = Gm1m2/r^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

What is the role of distance function in gravitational force in space exploration?

The distance function in gravitational force is essential in space exploration as it allows scientists to understand the gravitational interactions between planets, moons, and other celestial bodies. It is also used to calculate the trajectories of spacecraft and satellites.

How does the distance function in gravitational force change with distance?

The distance function in gravitational force follows an inverse square relationship, which means that as the distance between two objects increases, the force between them decreases exponentially. This means that the force between objects decreases quickly as they move further apart.

Are there any other factors that affect the distance function in gravitational force?

Yes, there are other factors that can affect the distance function in gravitational force, such as the masses of the objects and the presence of other objects in the vicinity. Additionally, the distance function is only an approximation and may not be accurate for extremely large or small distances.

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