# Gravitation and Circular motion.

1. Jun 22, 2006

### gunblaze

Hi, i surely do appreciate some help here..

How is gravity on the earth surface related to the circular motion of the earth about its axis?

If i'm not wrong, my lecturer once taught me that i can actually equate both gravitational force and centripetal force together since they are both the same. (Gravitational force holds all objects on earth together and move them in a circular orbit which is also what centripetal force does on the objects moving in circular motion.) which i think by further simplifying the equation GM/r^2 = rw^2 gives kepler's law, T^2 is directly proportionate to r^3.

Alright, now here's the problem. If both GM/r^2 and rw^2 can be equated like what my lecturer said, then both of the equation should give the value of gravitational field strength of the Earth on its surface which is (9.81ms^-2) However, by workin out rw^2, i'm unable to get the value 9.81ms^-2, what i get was only 0.03ms^-2. Is there a problem with my calculations or had i miss out certain things. I'm really quite confused over this.

Any help given will be greatly appreciated. Thanks!

2. Jun 22, 2006

### nrqed

If an object moves along a circle, then it is true that the net radial force will be equal to $m {v^2 \over r} = m r \omega^2$ and the acceleration is indeed [itex] r \omega^2 [/tex] . So far so good.

But how did you get 0.03 m/s^2? I am assuming that you used the period of rotation of the Earth an dthe Earth's radius? But notice that when you do that, you are considering an object sitting at the equator...resting on the ground. So the net force is NOT just the force of gravity! It is the gravitational force acting down AND the normal force acting up. The gravitational force is just a tiny bit larger than the normal force and that small excess is what leads to the centripetal force. So you cannot use GmM/r^2 for the total force on the object then!

If you use GmM/r^2 for the only force, then what you are considering is an object in *orbit* near the surface of the Earth. *Then* the acceleration will be 9.8 m/s^2 but the object will be moving much faster than the surface of the planet! It will be zooming by at a very large speed (so you cannot use the period of rotation of the Earth to find omega. Omega will come out whatever value it must be to give an acceleration of 9.8 m/s^2).

Patrick

3. Jun 23, 2006

### andrevdh

Things are a little more complicated than that (they usually are)...

The little stickman is holding a pendulum in his hand. The pendulum is hanging straight down (for him). We know it is experiencing two real forces - the weight $$W$$ and the tension $$T$$ in the string. We also know the pendulum is rotating around the earth's axis, so these two force combined should produce the required centripetal force:

$$\vec{F_C} = \vec{W} + \vec{T}$$

this explains why the pendulum is not hanging straight down towards the centre of the earth but at some odd angle.

Last edited: Nov 29, 2006
4. Jun 24, 2006

### andrevdh

We concluded that N2 requires that the pendulum should hang at some angle w.r.t. the local vertical such that [attachment 1]

$$\vec{F_c} = \vec{W} + \vec{T}$$

If he were to let go of the pendulum it will fall "straigh down" along the direction of the string of the pendulum. He would therefore conclude that the weight of the pendulum is acting along the direction of the string. If he were to measure its weight with a spring balance the pendulum would pull on it with a force equal to $$T$$. So we conclude that the apparent weight of the pendulum in his reference frame, $$W_a$$, is given by

$$\vec{W_a} = -\vec{T}$$

which according to the first equation comes to

$$\vec{W_a} = \vec{W} - \vec{F_}c$$

therefore

$$m\vec{g}_a = m\vec{g} - ma_c\hat{r}$$

where $$\hat{r}$$ is a unit vector pointing along $$r$$ from the axis to our little stickman.

$$\vec{g}_a = \vec{g} - a_c\hat{r}$$

the centripetal acceleration therefore modifies the local gravitational acceleration [attachment 2].

Last edited: Nov 29, 2006