# Gravity and Distance

1. Apr 8, 2007

### Izzhov

Newton's Law of Universal Gravitation states that: $$F_g = G \frac{m_1 m_2}{r^2}$$. My question is this: How can this law be used to derive a function for the change in distance between two masses ($$m_1$$ and $$m_2$$) a distance r apart as a function of time?

2. Apr 8, 2007

### quasar987

Well, Newton's Law of Universal Gravitation, no matter how many capitals you put in there, is pretty useless when not accompanied by Newton's Second Law of Motion:

$$\vec{F}=m\frac{d^2\vec{r}}{dt^2}$$

Actually, two equations of motion are necessary for this problem: one of each mass.

$$-G\frac{m_1 m_2}{|\vec{r}_1-\vec{r}_2|^3}(\vec{r}_1-\vec{r}_2)=m_1\frac{d^2\vec{r}_1}{dt^2}$$

$$G\frac{m_1 m_2}{|\vec{r}_1-\vec{r}_2|^3}(\vec{r}_1-\vec{r}_2)=m_2\frac{d^2\vec{r}_2}{dt^2}$$

These are two coupled differential equations, but see what you get by substracting the second fromt he first: One ordinary differential equation for $\vec{r}$, the vector going from m1 to m2. Its norm is precisely r, the distance btw m1 and m2, and solving the equation tells you how r varies with time.

Last edited: Apr 8, 2007
3. Apr 8, 2007

### arildno

You have forgotten unit vectors, dearest!

4. Apr 8, 2007

### Izzhov

Brilliant! There are only two problems: I have no idea how to solve differential equations, and I don't really have any idea what you're talking about.

Do you think you could just tell me the answer?

Last edited: Apr 8, 2007
5. Apr 8, 2007

### quasar987

Well I've answered your question from post #1, now you want me to solve the problem and tell you the answer? What is this about anyway? I suppose it's not for a homework, so why do you want to know?

It's complicated, but it turns out that there are 3 possible motions for m1 around m2 depending on the total energy and angular momentum: an ellipse (of which a circle is a special case), a parabola and an hyperbola. You can find plenty of information on this on the web. See for instance http://en.wikipedia.org/wiki/Two-body_problem#Application_to_inverse-square_force_laws.

Last edited: Apr 8, 2007
6. Apr 9, 2007

### Izzhov

First of all, this isn't a homework; it's a question I thought of in my free time (I'm in 9th grade; I don't think you learn differential equations until you're slightly older than that). Secondly, I didn't mean the circular/elliptical motion of a mass moving with a tangential speed around another mass, I meant the distance covered by two masses, just sitting there, motionless, in space, when the only force affecting them is their gravitational force on each other.

7. Apr 9, 2007

### arildno

Okay, so you are basically wanting to use the equation to derive when they hit each other?

8. Apr 9, 2007

### HallsofIvy

Staff Emeritus
Unfortunately, the answer to your original question is "solve the differential equation you get by putting the gravitational force into
$$F= ma= \frac{d^2x}{dt^2}. The problem simply cannot be done without at least some calculus. In fact, that's why calculus had to be invented! 9. Apr 9, 2007 ### arildno Newton used Euclid in Principia, I think. (But he probably "cheated" in his work room using his fluxions instead..) 10. Apr 9, 2007 ### Izzhov Kind of, except I want it to be assumed that both bodies are starting at rest, and that the only force acting on them is their gravitational force on each other. Also, I want the equation to tell how much distance is covered between them over a time t, not just when they collide. Well, I do know some calculus, just not differential equations or vectors. All I really know are the basic principles of differentiation and integration. Is there any chance it can be put in a way I would understand, with only some basic knowledge of calculus? 11. Apr 12, 2007 ### Izzhov I think I've found the solution to my problem. In order to know whether it's right, however, I need to know whether this is true: [tex] \frac{ \partial v(d)}{ \partial d} = \frac{1}{t}$$

In this equation, v(d) is velocity as a function of distance, d is distance, and t is time. So, is this equation true?

Last edited: Apr 12, 2007