Solución

Para demostrar que F\bold{F} no tiene espín (giro), calculamos su rotacional. Deja que P(x,y,z)=x(x2+y2+z2)3/2,Q(x,y,z)=y(x2+y2+z2)3/2P(x,y,z) = \frac{x}{\big(x^2+y^2+z^2\big)^{3/2}}, Q(x,y,z) = \frac{y}{\big(x^2+y^2+z^2\big)^{3/2}} y R(x,y,z)=z(x2+y2+z2)3/2R(x,y,z) = \frac{z}{\big(x^2+y^2+z^2\big)^{3/2}}. Entonces,

rot  F=−Gm1m2[(Ry−Qz)i+(Pz−Rx)j+(Qx−Py)k]=−Gm1m2[(−3yz(x2+y2+z2)5/2−−3yz(x2+y2+z2)5/2)i        +(−3xz(x2+y2+z2)5/2−−3xz(x2+y2+z2)5/2)j        +(−3xy(x2+y2+z2)5/2−−3xy(x2+y2+z2)5/2)k]=0\begin{aligned} rot\;\bold{F} &= -Gm_1m_2\big[(R_y - Q_z)\bold{i} + (P_z - R_x)\bold{j} + (Q_x - P_y)\bold{k}\big]\\ &= -Gm_1m_2\bigg[\bigg(\frac{-3yz}{\big(x^2+y^2+z^2\big)^{5/2}} - \frac{-3yz}{\big(x^2+y^2+z^2\big)^{5/2}}\bigg)\bold{i}\\ &\;\;\;\;+ \bigg(\frac{-3xz}{\big(x^2+y^2+z^2\big)^{5/2}} - \frac{-3xz}{\big(x^2+y^2+z^2\big)^{5/2}}\bigg)\bold{j}\\ &\;\;\;\;+ \bigg(\frac{-3xy}{\big(x^2+y^2+z^2\big)^{5/2}} - \frac{-3xy}{\big(x^2+y^2+z^2\big)^{5/2}}\bigg)\bold{k}\bigg]\\ &= 0 \end{aligned}

Dado que el rotacional del campo gravitacional es cero, el campo no tiene giro.