# Experiment to find the Center of Gravity under non-uniform $g$-field

When asked about how to find the center of gravity (CG) for a system, the answer normally goes like "suspend the system at different points and see how the vertical lines pass through. The CG is the intersection point of the vertical lines".

Does this suspension method still work in non-uniform gravitational field in determining the location of CG?

For instance, consider a system of particles of different masses joined by light, long rigid rods. If now the system is taken out from the Earth, the particle closer to the Earth experiences a larger g-field than the one furthest away. How can its CG be found experimentally?

• Possible duplicate by OP: physics.stackexchange.com/q/735469/2451 Commented Nov 6, 2022 at 14:28
• Hi Kim Leung. Welcome to Phys.SE. Please don't repost a closed question in a new entry. Instead, you are supposed to edit the original question within the original entry. Commented Nov 6, 2022 at 14:29
• perhaps this equation can help you ? \begin{align*} &R_{CM}=\frac{\sum_{i=1}^{3}\,m_i\,g(h_i)\,r_i}{\sum_{i=1}^{3}\,m_i\,g(h_i)} \quad\text{where}\\ &g(h)=\frac{M_E\,G}{(R_E+h)^2} \end{align*} and $~r_i\approx\,h_i$
– Eli
Commented Nov 6, 2022 at 17:56

I'd say that you want to evaluate the equivalent load distribution, namely

• the resultant of the force field and
• its line of application.

The result is likely to depend on the relative orientation of the system and the volume force field. We can introduce two sets of coordinates, one fixed with the force field, $$\mathbf{r}$$, one fixed with the system, $$\mathbf{r}^0$$. These two sets of coordinates may be related by translation of the origins and the rotation,

$$\mathbf{r}_P(\mathbf{r}_P^0; \mathbf{r}_O, \theta) = \mathbf{r}_O + \mathbb{R}(\theta) \cdot \mathbf{r}^0_P$$.

where $$\theta$$ are the parameters describing the rotation of the system around the origin of the reference $$^0$$, and $$\mathbf{r}_O$$ is the relative position of the origin of the reference $$^0$$, from which the resultant and the "center of the force" will depend on.

• The resultant of the force reads

$$\displaystyle \mathbf{F}(\mathbf{r}_O, \theta) = \int_{V^0} \rho(\mathbf{r}^0) \mathbf{g}(\mathbf{r}(\mathbf{r}^0; \mathbf{r}_O, \theta)) dV^0$$

• The application line for the configuration is defined as the line along which the resultant needs to act on the system to have an equivalence to the moments

$$\displaystyle \mathbf{r}_H \times \mathbf{F}(\mathbf{r}_O, \theta) = \int_{V^0} \rho(\mathbf{r}^0) \ \mathbf{r}(\mathbf{r}^0; \mathbf{r}_O, \theta) \times \mathbf{g}(\mathbf{r}(\mathbf{r}^0; \mathbf{r}_O, \theta)) dV^0$$.

It should not be difficult to see that this relation doesn't provide a single point, but a family of point, the application line, since $$(\mathbf{r}_H + s \mathbf{\hat{F}}(\mathbf{r}_O, \theta) ) \times \mathbf{F}(\mathbf{r}_O, \theta) = \mathbf{r}_H \times \mathbf{F}(\mathbf{r}_O, \theta)$$, for every vector $$s \mathbf{\hat{F}}(\mathbf{r}_O, \theta)$$ aligned with the resultant force $$\mathbf{F}(\mathbf{r}_O, \theta)$$, since $$\mathbf{\hat{F}}(\mathbf{r}_O, \theta) \times \mathbf{F}(\mathbf{r}_O, \theta) = \mathbf{0}$$