# Variation of gravity with latitude

1. Sep 4, 2006

### physics_lover

When an observer stand on at the place with a angle "a" of the latitude,
the observed value of the gravity will be affected by the motion of the rotation.

SO what is the value of the obversed gravity at that time???:surprised

Last edited by a moderator: Sep 15, 2015
2. Sep 4, 2006

### Kurdt

Staff Emeritus
The force of gravity is counteracted by the centrifugal force so the observed value of gravity is less than the true value. The effect due to the ficticious centrifugal or centripetal force (however you choose to name it) is as follows.

$$\mathbf{F}_c = m(\mathbf{\omega} \times (\mathbf{\omega} \times \mathbf{r}))$$

Where $$\mathbf{\omega}$$ and $$\mathbf{r}$$ are the angular velocity and radius of the earth respectively.

Last edited: Sep 5, 2006
3. Sep 5, 2006

### physics_lover

thz very much, the obversed gravity g' is less than ture value of g =10ms^-2 .
But I can't find the equation of g' in term of g & "angle of a" & "r"
& angluar velocity W only,

help me again pls!!! :surprised

4. Sep 5, 2006

### Kurdt

Staff Emeritus
You need to resolve the centripetal acceleration into horizontal and vertical components (think about what would be the best definition of horizontal and vertical in this case).

5. Feb 28, 2010

### Charlie Costa

But at a latitude of $$\varphi$$, isn't the radius in the above expression not the radius of the earth, r, but rcos$$\varphi$$? Assuming of course a spherical earth, which is not really correct.