# Distance function in gravitational force

1. Oct 20, 2009

### coki2000

Hi,
In gravitational force law $$F=-G\frac{M_{1}M_{2}}{x^{2}}$$if i want to find the distance function which depends on time then

$$\frac{d^{2}x}{dt^{2}}=-G\frac{M}{x^2}$$

$$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-G\frac{M}{x^2}$$

$$vdv=-G\frac{M}{x^2}dx$$

integrate the both sides

$$\frac{v^2}{2}=G\frac{M}{x}\Rightarrow \frac{dx}{dt}=\sqrt{2G\frac{M}{x}}\Rightarrow dt=\sqrt{\frac{x}{2Gm}}dx$$

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

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

Last edited: Oct 21, 2009
2. Oct 21, 2009

### pc2-brazil

Good morning,

Your TeX formulas are not being displayed in image form.
You have to put $$$$$$$$ tags around the TeX formulas so that they are displayed correctly. Like this:
$$\textbf{$$}$$F=-G\frac{M_{1}M_{2}}{x^{2}}$$.
I don't know how to answer your question, but I will follow this post, because I'm also very interested in knowing.

3. Oct 24, 2009

### pc2-brazil

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:
$$\frac{v^2}{2}+C_1=G\frac{M}{x}+C_2\Rightarrow \frac{dx}{dt}=\sqrt{2G\frac{M}{x}+C}\Rightarrow dt=\sqrt{\frac{x}{2Gm+Cx}}dx$$
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.

4. Oct 24, 2009

### Cleonis

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.

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

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.

$$\ddot r = -G M r^{-2} + \dot\theta^2 r$$

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 $$\dot\theta r^2$$

$$\dot\theta r^2 = constant \qquad \Rightarrow \qquad \frac{d(\dot\theta r^2)}{dt} = 0$$

Differentiating, using the product rule:

$$r^2 \ddot\theta + \dot\theta \frac{d(r^2)}{dt} = 0$$

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

$$r^2 \ddot\theta + 2 r \dot\theta \dot r = 0$$

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

$$\ddot\theta = - \frac{2 \dot\theta \dot r}{r}$$

Notated as a system of equations:

$$\ddot r = -G M r^{-2} + \dot\theta^2 r$$

$$\ddot\theta = - \frac{2 \dot\theta \dot r}{r}$$

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

Cleonis

Last edited: Oct 24, 2009
5. Oct 24, 2009

### Cleonis

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. Oct 24, 2009

### ideasrule

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. Oct 25, 2009

### coki2000

In my opinion, I don't forget the constant;

$$vdv=-G\frac{M}{x^2}dx$$ integrating both sides

$$\frac{v^2}{2}=G\frac{M}{x}+C$$

But multiply the both sides by m

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

$$\frac{1}{2}mv^2=E_k=\lim_{x\rightarrow \propto }{G\frac{Mm}{x}}+C=0$$

So when the distance of the two particles equal to infinity, kinetic energy is zero.So C=0.

8. Oct 25, 2009

### ideasrule

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.

9. Oct 25, 2009

### D H

Staff Emeritus
You don't know that. You are arriving at C=0 by assuming that $$v_{\infty} = 0$$, and that is not a valid assumption.

10. Oct 25, 2009

### coki2000

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 $${C}\neq0$$ then kinetic energy$${E_k}\neq{0}$$ when distance is infinity.If $${C}\neq0$$ then the formula is wrong.So it may be zero.

Last edited: Oct 25, 2009
11. Oct 25, 2009

### D H

Staff Emeritus
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. Oct 27, 2009

### coki2000

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.