# Accleration due to gavity

1. Aug 8, 2011

### arul_k

Is it possible that for two very large objects (like the earth and the moon), brought close together, the resultant accleration coud be the combination of both their gravitational fields?

We know for less massive objects accleration due to gravity (on Earth) is apparently constant at 9.8m/s^2. Could this be due to the fact that their gravitational fields are weak and this may not hold true if the objects are very large.

2. Aug 8, 2011

### ZapperZ

Staff Emeritus
Start by reading the FAQ subforum in the General Physics forum.

Secondly, you said "the resultant acceleration"... but you never specify the resultant acceleration of which body!

Zz.

3. Aug 8, 2011

### arul_k

The FAQ on accleration ends like this:

However, if we change that situation, i.e. if m1 is now comparable, or even bigger, than M, then those simplifications will no longer be valid, and one will have to start from the very beginning to figure this out.

May be you could explain further.

As far as which body undergoes the accleration, since both the bodies are massive, I assume both would undergo accleration

4. Aug 8, 2011

### HallsofIvy

Staff Emeritus
Because you haven't said what "simplification" this is talking about, this fragment doesn't mean anything.

Strictly speaking this is corrrect since "massive" could simply mean "has mass". But I don't think that's what you mean. For any two objects that have mass, no matter how small one or both are, each object attracts the other and both objects accelerate.

$$F= \frac{GmM}{r^2}$$
is symmetric. The moon and the earth, or the earth and a rock, both attract each other exactly the same force. Of course, the acceleration is that force divided by the mass of the object so the acceleration of the moon is much larger than that of the earth and the acceleration of the rock is much greater than that of the earth.

At the surface of the earth, G, M (the mass of the earth), and the r (the distance to the center of the earth) are constant so from
$$F= \frac{GmM}{r^2}$$
and F= ma, so that a= F/m,

we get, for any object, no matter what its mass,
$$a= \frac{F}{m}= \frac{GM}{r^2}$$
is the same.

5. Aug 8, 2011

### m.e.t.a.

If I understand your original question correctly, then HallsofIvy has stated the equation necessary to find the resultant acceleration. It might help to define "resultant acceleration", $a_{\rm{res}}$, as $\ddot r$, where $r$ is the distance between the two bodies. So $a_{\rm{res}}$ is not the acceleration of body 1, nor the acceleration of body 2, but the sum of the (absolute values of the) accelerations of the two bodies:
$${a_{{\rm{total}}}} = \left| {{a_1}} \right| + \left| {{a_2}} \right|$$

6. Aug 8, 2011

### arul_k

.

You an view the entire post in the FAQ section.

I agree, I didn't mean to ask about a comparision of accleration of the objects involved.

This is exactly my question, the second mass no matter how large, it may be even twice the mass of the earth, will acclerate at the same rate (9.8m/s^2 in case of the earth). why dosen't the larger gravitional feild of the second mass (twice that of the earth) produce a larger accleration?

7. Aug 8, 2011

### arul_k

Yes, this is what I mean by resultant accleration.

8. Aug 8, 2011

### m.e.t.a.

Going back to Newton's law of gravitation:
$${\vec F_{12}} = - \frac{{G{m_1}{m_2}}}{{{{\left| {{{\vec r}_{12}}} \right|}^2}}}{\hat r_{12}}$$
This equation, which provides a highly accurate description of gravity, states that the gravitational forces, ${\vec F_{12}}$ and ${\vec F_{21}}$, experienced by $m_1$ and $m_2$ respectively, are equal and opposite. HallsofIvy has also pointed out that the force experienced by $m_1$ due to the gravitational field of $m_2$ is proportional to the mass of $m_1$, and vice versa. As has been already shown in this thread, this explains why heavy objects don't fall faster than light objects. If you understand this, then you will notice that Newton's equation sets no particular limits on the magnitudes of $m_1$ and $m_2$. The equation holds equally true for objects of 1 kg as it does for objects the mass of the Earth and beyond.

Hence, for given values of $m_1$ and $r$, an object $m_2$ will experience exactly the same gravitational acceleration towards $m_1$, regardless of its mass.

9. Aug 8, 2011

### mikeph

Because it doesn't accelerate under its own gravitational field.

The distinction you're looking for is one between inertial mass (relationship governing the acceleration a body undergoes when subjected to a force, a = F/m_intertial) and gravitational mass (the m term in Newton's equation, the force of interaction being GMm_gravitational/r^2). If inertial mass = gravitational mass, then the two quantities will cancel, a = F/m = GMm/mr^2 = GM/r^2.

People are still doing experiments to determine whether inertial mass is the same as gravitational mass, and after 100s of years of testing, nobody has yet disproved it. So acceleration purely due to the gravitational force is not a function of a body's mass.

10. Aug 8, 2011

### Staff: Mentor

I think the more massive object accelerates the Earth quicker than a lower mass object. So an object's acceleration from an outside reference point is equal no matter its mass, the Earth is accelerated at different amounts depending on the mass of the other object. The speed at which the 2 objects would close would increase as one of the objects mass increases. At least I think that is the case.

11. Aug 8, 2011

### TurtleMeister

Acceleration is frame dependent.

In the two body problem, the inertial frame of reference is the common center of mass of the two bodies. It is in this frame that the universality of free fall applies. And it is true regardless of the difference in mass of the two bodies. The acceleration of m1 is not affected by it's own mass, and likewise, the acceleration of m2 is not affected by it's own mass.

$$A_{m1}=\frac{G{m2}}{R^2}$$
$$A_{m2}=-\frac{G{m1}}{R^2}$$

If the frame of reference is either one of the two bodies, then it is called the relative acceleration. This frame is non-inertial, so the universality of free fall does not apply. It is the sum of the absolute accelerations of m1 and m2. It is the acceleration of one body relative to the other.

$$A_{rel}=\frac{G({m1}+{m2})}{R^2}=\left | A_{m1} \right | + \left | A_{m2} \right |$$

In both scenarios the time to impact will be the same, but the accelerations depend on the frame of reference.

Last edited: Aug 8, 2011
12. Aug 9, 2011

### arul_k

As you stated when you combine the 2 equations you get a = GM/r^2 and you always get a constant accleration because the second mass becomes irrelevent

In my personal view this would only hold true for masses much smaller than the first (larger) mass. It seems logical that if the second mass were large enough to produce a significant accleration on the first then the apparent accleration could not be constant.

13. Aug 9, 2011

### mikeph

Sorry, I mean constant for any mass you pick.

You're completely correct, if the acceleration is GM/r^2, then as a system evolves (under the influence of this acceleration), r^2 may change as well, leading to a different acceleration. It's a major un-simplification that you're discussing, which is why I'd like to avoid it unless it's crucial to answering your question. The general case is a two-body problem leading to elliptical, parabolic or hyperbolic trajectories.

14. Aug 9, 2011

### arul_k

The frame of reference does come into the picture as mentioned by TurtleMeister and Drakkith, but how do you come to the conclusion that from an external frame of reference (inertial frame) the accleration of the Earth would be constant? The resultant accleration could be due to the combined gravitational force that the two objects exert on each other (naturally both objects would experience the same amount of force).

15. Aug 9, 2011

### TurtleMeister

Your personal view goes against the universality of free fall. And then you ask...
I think it is your personal view that is holding you back. You need to let go of it long enough to understand how the UFF works. Once you understand it, I think you will find that your personal view is flawed. In fact, I may be able to convince you that there is a problem with your view.

You seem to think that the UFF is only true when one mass is much smaller than the other. Well then, where do you draw the line? I mean how much smaller? Is there a certain mass difference, or ratio, that must be reached where the UFF goes from being false to true? If so, how do you determine it?

16. Aug 9, 2011

### Staff: Mentor

If the acceleration of one body was the result of both its and the other objects gravity, then I don't think a more massive object would orbit at the same distance and velocity as a smaller body does with the exact same orbital properties. Meaning that if we replaced the Earth with the Moon, currently it would still occupy the same general orbit that the Earth does. In your view then the less mass of the Moon would mean that the force holding it in its orbit would be less and it would be lost.

17. Aug 10, 2011

### agentredlum

Suppose the distance between us is 10m, and suppose i have an accurate stopwatch. If you accelerate toward me at 1m/s^2 and i accelerate toward you at 1m/s^2 then a stationary observer would measure both our accelerations at 1m/s^2 in opossite directions such that a collision between us will occur. Now, if i consider you at rest then i conclude that my acceleration is 2m/s^2 because of the rate at which distance is changing between us. (2nd derivative of distance as a function of time) But that is not true because it would imply a force on me DOUBLE of what the actual force is because F = ma.

Assuming objects are stationary is dangerous because it leads to the wrong answers.

Assuming anything about objects is dangerous. To make the point, if i assume your acceleration is 8m/s^2 away from me then i must conclude that my acceleration is 10m/s^2 toward you. This is clearly wrong from the stationary observers point of view, who is the only accurate judge in this experiment, and makes me overestimate the force on me producing my acceleration by 10X.

So in order to make any accurate measurements we need stationary observers.

Last edited: Aug 10, 2011
18. Aug 10, 2011

### TurtleMeister

Hi agentredlum. I'm not disagreeing with you, and I think that your post is correct. However, I would just like to point out that in your scenario, the two bodies have two separate forces that can have different magnitudes. In the two body problem, the forces are always equal and opposite in accordance with the third law of motion. So if you know the relative acceleration and the mass of both bodies, then the correct force can be calculated without a stationary observer (inertial frame of reference).

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