# Force of Gravity

1. Dec 27, 2003

### napoleonmax

I was thinking about Newton's gravity equation:
F = G * m(1) *m(2) / r^2. I wanted to find the acceleartion due to gravitational attraction that would occur between two objects of different masses so I used the equation F=ma. Then I ran into difficulty: should I use the sum of the masses to solve this or should I use both masses individually. In other words should my acceleration equation be a= F / m(1) + F / m(2) OR
a = F / [m(1)+m(2)] ?????? Which one is correct, if any? Clearly the two are not equal (if m(1) is 4 and m(2) is 3 then by the first equation the acceleration will be F/4 + F/3 which is 7F/12 and by the second equation the acceleration is F/7). Another problem is that these equations would allow heavier things to fall faster to the Earth (though only significantly on a large scale). That is, the moon would fall faster toward the Earth than would a car because the change in the force would exede the change in the mass. I would appreciate any help on the matter.

2. Dec 27, 2003

### Staff: Mentor

F=ma is indeed symmetrical with respect to gravitational attraction. So when you jump out of a plane, the force accelerating you down toward the earth is also accelerating the earth up toward you.

And yes, a veeeeery large object would not accelerate at G - G works only for objects which are an insignificant fraction of the mass of the earth (ie, it works pretty well for anything this side of celestial mechanics).

3. Dec 28, 2003

### Staff: Mentor

Neither is correct. The right way is to use a1 = F/m1.
Using the correct equations, you'll find that acceleration is independent of mass. Thus, if they were at the same distance from the earth, the car and the moon would have the same acceleration. Does this help?

4. Dec 28, 2003

### Staff: Mentor

Re: Re: Force of Gravity

What if you put Jupiter at that distnce from the earth? Will the acceleration still be the same?

5. Dec 28, 2003

### Jimmy

Re: Re: Re: Force of Gravity

Oh no! Well, if Earth exerts the same gravitational force on Jupiter that Jupiter exerts on Earth...

6. Dec 28, 2003

Re: Re: Re: Force of Gravity

The acceleration of Jupiter towards the Earth would be the same as a car and the moon, yes. BUT, the acceleration of the Earth towards Jupiter would be substantially larger than than the acceleration of the Earth towards a car!!
Therefore, for an observer on Earth, Jupiter would APPEAR to accelerate faster.

7. Dec 28, 2003

### Jimmy

Re: Re: Re: Re: Force of Gravity

It seems counter-intuitive but it does indeed work out that way.

If you calculate the gravitational force between the Earth and Jupiter,

$$F_g = \frac{G M m}{r^2}$$

and then calculate Jupiter's Acceleration toward the earth,

$$a = \frac{F}{m}$$

it comes out the same as any other mass, large or small, at the same distance. That is what you are doing when you use the equation,

$$a_g = \frac{G M}{r^2}$$

After all,

$$m a = \frac{GMm}{r^2}$$

Last edited: Dec 28, 2003
8. Dec 29, 2003

### Arcon

I made a web page which gives a similar problem. I used two objects of the same mass though. It is not difficult to extend to different masses. Please see bottom of this web page

http://www.geocities.com/physics_world/gr/grav_field.htm

To find the force on an object simply use the principle of superpostition, i.e. add the forces on the "test" body which result from all the other bodies. You'll find that if the other bodies don't move then the acceleration on the test body is independant on the mass of the test body.

9. Dec 29, 2003

### Staff: Mentor

My point was just that the simplifications you can use for two objects of vastly different mass don't work so well with two objects of similar mass. Whether you drop a bowling ball or a car out of a plane, it works fine to consider the earth completely stationary (not accelerating toward the plane). But when you drop the moon out of that plane, you need to account for both accelerations and probably pick a better reference frame (at the center of mass of the system for example).

10. Dec 29, 2003

### Arcon

Yes. Excellant point of course. Funny that people rarely aks that question. I was posting a forum in the late 90's called Ask Dr. Neutrino and recall a person asking that very astute question and recall in discust that he was flamed for it.

Independance of acceleration in free-fall demands that the falling object be a test particle, i.e. that it is pointlike and has a mass which may be neglected in comparison to the source of the field. To obtain the answer that you seek you can simply replace the mass of the source with the sum of the masses of the two particles. I.e. suppose the two particles are constrained to move along the x-axis. Then it can be shown that

$$A = \frac {G(m_1 + m_2)} {R^2}$$

where A is the relative acceleration of the two particles and R is the distance between them.

http://www.geocities.com/physics_world/mech/two_accel.htm

Integral - Thanks for showing me that! :-)

11. Dec 29, 2003

### Arcon

Yes. Use the sum of the masses. But note that this applies to a special case of the particles moving along a line.

12. Dec 29, 2003

### Staff: Mentor

You are of course correct that when talking about the acceleration with respect to the earth of a massive body, then the acceleration of the earth must be factored in. I was taking an inertial frame as my reference point (silly me).

For straight line motion, it's trivial (as Arcon points out). My advice to napoleonmax is to keep it simple. Find the accelerations of each using an inertial frame (where F=ma), then add them to get the relative acceleration. This results in napoleonmax's first equation:

Yes. That gives the correct relative acceleration.

Edit: included quote by napoleonmax and comment.

Last edited: Dec 30, 2003
13. Dec 29, 2003

### Staff: Mentor

Its just that earth is the reference frame people are used to. When they get a situation where the earth isn't really able to be considered stationary, it throws them a curve. In some cases, earth can be considered an inertail frame, in some it can't.