# Free-fall acceleration vs gravitational acceleration

1. Dec 1, 2013

### Wenchao.Zhang

Hi, experts
I got a naive question about the relation between the free-fall acceleration vs gravitational acceleration.
When you consider the rotation of the earth around the axis which runs through the north and south poles, the free-fall acceleration "g" of an object with mass m is different with its gravitational acceleration "a_g".
For example, when the object is at the equator of the earth, then "m*g=m*a_g-m*w^2*R", where "w" is the angular velocity of the earth rotation.
This exmaple is a simple case. What I do not understand is the case that the object is located in a place which is between the equator and poles.
For this case, the component of the gravitational force which is perpendicular to rotational axis acts as the centripetal force for the rotation. How about the component of the gravitational force which is parallel to the rotational axis? Is the magnitude of this component equal to the weight of the object m*g? Or how to relate the free-fall acceleration "g" with the gravitational acceleration "a_g" for this case?
Best wishes
W.

2. Dec 1, 2013

### Astrum

Your post was difficult to follow, but I understand that you're confused about the "component of gravity" here.

When we say $g=9.8 \frac{m}{s^2}$, we're talking about the average acceleration due to gravity across the earth. The Earth is an oblate spheroid, which means that the distance from the center of the earth is slightly different at the equator versus the poles. This is what is reasonable for the difference in a local $g$.

I didn't understand what you were trying to say in the middle of your post. What is $a_g$ suppose to be?

3. Dec 1, 2013

### Wenchao.Zhang

Hi, Astrum
Here a_g means the gravitaional acceleration. The corresponding gravitational force is F=G*m*M/R^2, where M is the mass of the earth, m is the mass of the object, G is the gravitational constant. Thus a_g=G*M/R^2
If you do not consider the rotation of the earth, then m*g is equal to m*a_g. When you consider the rotation, m*g is not equal to m*a_g.
My question is what's the relation between a_g and g in a point which is located between the equator and poles?
W.

Last edited: Dec 1, 2013
4. Dec 1, 2013

### Astrum

Oh, I see what you're asking about. What you'd need to find is the centripetal acceleration at the surface of the earth and hence, $\mathbf v$.

Use $\mathbf v = \mathbf \omega \times \mathbf r$ to find this, then you can plug it into the equation for radial acceleration.

Last edited: Dec 1, 2013
5. Dec 1, 2013

### Wenchao.Zhang

The equation "m*g=m*a_g-m*w^2*R" is only true for the case that the object is on the equator. For this case, what I need is to find the centripetal acceleration "w^2*R". When the object is not on the equator (e.g. the object is on the surface of the earth at north latitude of thirty degrees), "m*g=m*a_g-m*w^2*R" does not hold any more. We need a new equation for the relation between a_g and g.
W.

Last edited: Dec 1, 2013
6. Dec 1, 2013

### Astrum

What? You need to find the centripetal acceleration, yeah. What you have written $ma_g-\omega^2 R$ is, as you say, for when $\theta = \pi /2$.

Generalize everything, and you'll have your answer. $a=a_g + \frac{\mathbf F _r }{m}= a_g - \frac{(\mathbf \omega \times \mathbf r )^2}{r}$. The Earth's angular velocity is about $7.29 \times 10^{-5} rad/s$, plug that in and you'll see that value is extremely small.

7. Dec 1, 2013

### sophiecentaur

Reading between the lines here, I suspect you are assuming that a falling body is somehow affect by the spinning of the Earth. In fact, strictly it is not. The only centripetal force on the falling body is due to the gravitational field (when observed from an external frame of reference. However, if you were to fall down a vertical mineshaft, anywhere but at the Poles, you would find yourself hitting the wall of the shaft. You would 'be aware of' a force, which is called the Coriolis force and it is a force that is observed when you are in a rotating frame of reference and you try to move radially. It would have the effect of pushing you against the wall. It's explained by the fact that the peripheral speed of the wall lower down is less than the peripheral speed higher up (same angular velocity but smaller radius). The Coriolis force is only there for moving objects. It explains the movement of the wind in a curve around areas of high and low pressure, rather than moving radially, in and out,as one would expect at first sight.

The effect at the Equator is maximum and it gets less as you approach the Poles, as the radius of the circular motion, about the NS axis, gets less.

There is a secondary effect, due to the oblate shape of the Earth, because, at the Poles, you are closer to the Centre of Mass. Also, the mass distribution is not uniform inside the Earth. But it's best to ignore all that until the basic 'ideal' model is understood.

8. Dec 1, 2013

### xAxis

You are trying to calculate the acceleration of free falling body relative to a fixed point on earth under it which moves as the earth rotates? That is not very useful because the earth moves relatively slowly, and for the time when g in your equation is considered constant, it is virtually zero.