Torque Revisited — Question 6

Question 6

A gravity gradient produces torque about the mass center

Two equal masses $m$ are fixed at $\vec r_\pm=\pm a(\cos\theta,\sin\theta,0)$ relative to their midpoint. The external field is $\vec g(x)=-(g_0+kx)\hat y$, with $k$ in $\mathrm{s^{-2}}$.

1. Find torque about the midpoint.

2. Evaluate its sign at $\theta=0$ for $k>0$ and explain why the uniform-field shortcut fails.

Question 6 - Solution

A gravity gradient produces torque about the mass center

(a) Derivation and calculation

The horizontal coordinates are $x_\pm=\pm a\cos\theta$. Summing the two moments, $$\tau_z=-m\sum_\pm x_\pm(g_0+kx_\pm).$$ Since $x_++x_-=0$ and $x_+^2+x_-^2=2a^2\cos^2\theta$, $$\boxed{\tau_z=-2mka^2\cos^2\theta}.$$

(b) Result and interpretation

For $\theta=0$ and $k>0$, $$\boxed{\tau_z=-2mka^2<0}\quad\text{(clockwise)}.$$ The right mass feels a greater downward force. A single force $M\vec g(0)$ at the midpoint would have zero moment and omit this field-gradient torque; no common $\vec g$ can be factored out of the particle sum.

Physical check

The gradient moment vanishes at $\theta=\pi/2$, where both particles lie on the same vertical force line.