Acceleration due to gravity field

[Symbreak]

It is often said that a body will accelerate at the same rate in a gravitational field, independent of its mass.
But if every mass generates a gravitational field of its own, would not this add to the field in which it is falling - so a significantly large mass would fall at a faster rate?
If the gravitational acceleration (a) is the result of the gravitational field, it seems that a will be effected by the mass of any body. Maybe someone could help clarify this issue.

 

[russ_watters]

Actually, you can calculate separately the accelerations of the two bodies. Ie, the Earth is falling toward the moon and the moon is falling toward the earth, with two different accelerations when viewed by an observer who is stationary relative to our system's center of mass.

 

[quasar987]

In the Newtonian theory of gravitation (and there's probably an equivalent statement in Einstein's theory), the gravitational field produced by a body assumes a value everywhere except at the location of the body itself. In other words, the body does not influences itself gravitationally.

Suppose you have a particle of mass m. Then the grav. field it produces it

$$\vec{g}(r) = -\frac{Gm}{r^2}}\hat{r}$$

Notice how for r = 0 (i.e. the location of the particle itself), the field is not defined (because mathematically, the algebraic operation of dividing by 0 makes no sense).

 

[Janus]

There are two separate issues here. The acceleration a given body will have due to gravitation, and the speed it will have relative to the Body towards which it is "falling".

The force of gravity is porportional to the product of both masses, so both bodies contribute to the force pulling them together.

$$F_g = \frac{GMm}{d^2}$$

If m is the mass of the body for which we are considering at the moment then the acceleration will be F = ma

thus
$$ma = \frac{GMm}{d^2}$$

The "m"s cancel out and we get.
$$a = \frac{GM}{d^2}$$

no matter what the value of m, and this is the acceleration of m due to gravity

If we do the same for M we the same result; an acceleration for M that does not change with the value of M.

In other words, M falls towards m and m falls towards M. The rate at which they close distance is the sum of these two velocities.

If one of these masses increases mass, the rate at which it falls does not change but the rate at which the other mass falls will.

So if one of the masses were very large compared to the other, say the Earth and A ball, the amount of acceleration of the larger mass will be very, very small, and we usually can ignore it. Even if we double the mass of ball, the change in acceleration of the Earth is too small to notice. We would have to increase the ball to a significant percentage of the Earth's mass before we noticed any change, and the ball seemed to fall faster.

 

[Integral]

On the surface of the Earth we can use the common expression for gravitational force:

$$F = \frac {G M_e m} {R^2_e}$$

Where G is the Gravitational constant, $R_e$ is the radius of the earth, $M_e$ is the mass of the Earth and $m$ is the mass of some body near the surface of the earth.
Newton taught us that when a force is applied to a mass it accelerates like $F= m a$

Now since we have an expression for the gravitational force on a body of mass m, we can find its acceleration due to this force by combining these relationships.

$$F= ma = \frac {G M_e m} {R_e^2}$$

notice that the mass of the body in question appears on both sides of this expression so it cancels out leaving

$$a = \frac {G M_e } {R_e^2}$$

With a bit of googleing you ought to be able to find values for all the constants in the above expression. This will yield the commonly accepted value of ~9.8 $\frac m {s^2}$

Notice, also, that the mass of BOTH the Earth and the falling body are involved in generating this constant acceleration. So your concern has been considered to arrive at the commonly accepted conclusion that the acceleration due to gravity is constant near the surface of the earth. Note also that if you get a significant distant from the Earth's surface the acceleration will depend on that distance.

 

[Symbreak]

Thanks for clearing this issue up! I was somewhat confused over the 'addition of fields' in EM, but this is obviously not the case with gravity.
Cheers.

 

[Enos]

I've read that there is also air resistance, so that when the resistance equals the mass the falling object, it will no longer accelerate and stay at a uniform speed.

 

[quasar987]

Thanks for clearing this issue up! I was somewhat confused over the 'addition of fields' in EM, but this is obviously not the case with gravity.
Cheers.

It's the same thing for an electric field or a magnetic field. The field producer can at no time be influenced by its own field.

 

[tony873004]

...
$$a = \frac {G M_e } {R_e}^2$$
...

Should be:
$$a = \frac {G M_e } {R_e^2}$$
as well as the equations that led up to this one.

You got it right, but it was a TEX error.

 

[Grogs]

I've read that there is also air resistance, so that when the resistance equals the mass the falling object, it will no longer accelerate and stay at a uniform speed.

There certainly is (on Earth at least), but it has nothing to do with the gravitational acceleration, which is to say that while it has a retarding effect on falling objects and ultimately limits the velocity they reach, it doesn't change the way gravity works. Because accounting for air resistance gets to be pretty challenging mathematically, most low level physics problems will ignore air resistance. On real world problems like ballistics and rocket launches, it has to be accounted for.