A system of particles with masses \(m_{\alpha}\) moves in a gravitational field \(\boldsymbol{g}\). Show that if \(\boldsymbol{F}_{\alpha}=-m_{\alpha} \boldsymbol{g}\) where \(\boldsymbol{g}\) is constant, then \(\sum_{\alpha} \boldsymbol{F}_{\alpha}=m \boldsymbol{g} \quad\) and \(\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha}=m \boldsymbol{c} \wedge \boldsymbol{g}\), where \(c\) is the position vector of the centre of mass and \(m\) is the total mass. Deduce that the effect of a uniform gravitational fieldon the total linear momentum and angular momentum is the same as that of a single force \(m g\) acting through the centre of mass. Show by counter-example that this is not true for a non-uniform field.

We are given that \(\boldsymbol{F}_{\alpha} = -m_{\alpha}\boldsymbol{g}\), so we can first calculate the sum of forces, \(\sum_{\alpha} \boldsymbol{F}_{\alpha}\): $$\sum_{\alpha} \boldsymbol{F}_{\alpha} = \sum_{\alpha} (-m_{\alpha}\boldsymbol{g}) = -\boldsymbol{g} \sum_{\alpha} m_{\alpha}$$ Recall that the total mass of the system, \(m\), is just the sum of the individual particle masses, so \(m = \sum_{\alpha} m_{\alpha}\). Thus, we can substitute it into the above equation: $$\sum_{\alpha} \boldsymbol{F}_{\alpha} = -mg$$

Recall that the position vector of the centre of mass, \(\boldsymbol{c}\), is given by: $$\boldsymbol{c} = \frac{1}{m}\sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Now, let's calculate the sum of angular momenta, \(\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha}\): $$\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha} = \sum_{\alpha} (\boldsymbol{r}_{\alpha} \wedge (-m_{\alpha}\boldsymbol{g})) = -\sum_{\alpha} m_{\alpha} (\boldsymbol{r}_{\alpha} \wedge \boldsymbol{g})$$ Now, we can factor out the constant \(\boldsymbol{g}\): $$-\sum_{\alpha} m_{\alpha} (\boldsymbol{r}_{\alpha} \wedge \boldsymbol{g}) = -\boldsymbol{g}\sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Recall the equation for the centre of mass from before: $$\boldsymbol{c} = \frac{1}{m}\sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Multiplying both sides by the total mass \(m\), we get: $$m\boldsymbol{c} = \sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Substitute this into our equation for the sum of angular momenta: $$\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha} = -\boldsymbol{g}(m\boldsymbol{c})$$ So, $$\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha} = m \boldsymbol{c} \wedge \boldsymbol{g}$$

From the previous steps, we have shown that the sum of forces equals \(-mg\) and the sum of angular momenta equals \(m \boldsymbol{c} \wedge \boldsymbol{g}\). This means that the effect of the uniform gravitational field on the total linear momentum and total angular momentum of the system of particles is the same as that of a single force, \(mg\), acting through the centre of mass.

To show that the same relationships do not hold true for a non-uniform field, let's consider a simple system with two particles, each particle having a mass equal to \(m_{1} = m_{2} = 1\) kg, and located at distances \(r_{1} = 0\) and \(r_{2} = 1\) m, respectively, from the origin. In a non-uniform field, the gravitational force acting on each particle can be represented by \(\boldsymbol{F}_{1} = -k(r_1) \hat{\boldsymbol{r}}_1 = -0\hat{\boldsymbol{r}}_1\) and \(\boldsymbol{F}_{2} = -k(r_2) \hat{\boldsymbol{r}}_2 = -k\hat{\boldsymbol{r}}_2\), where \(k\) is a constant. Thus, the sum of forces is: $$\boldsymbol{F}_{1} + \boldsymbol{F}_{2} = 0 - k\hat{\boldsymbol{r}}_2 \neq -mg$$ In this case, the sum of angular momenta is: $$\boldsymbol{r}_{1} \wedge \boldsymbol{F}_{1} + \boldsymbol{r}_{2} \wedge \boldsymbol{F}_{2} = \boldsymbol{r}_{1} \wedge 0 + \boldsymbol{r}_{2} \wedge (-k\hat{\boldsymbol{r}}_2) = -r_2 k \hat{\boldsymbol{r}}_3 \neq m\boldsymbol{c} \wedge \boldsymbol{g}$$ Therefore, for a non-uniform gravitational field, the total linear momentum and the total angular momentum relationships do not hold true, as shown by this counter-example.