SUPERPOSITION OF GRAVITATIONAL FORCES
Many stars belong to systems of two or more stars held together by their mutual gravitational attraction. Figure 13.5 shows a three-star system at an instant when the stars are at the vertices of a 45° right triangle. Find the total gravitational force exerted on the small star by the two large ones.
IDENTIFY, SET UP, and EXECUTE:
We use the principle of superposition: The total force \vec{w} on the small star is the vector sum of the forces \vec{w}_1 and \vec{w}_2 due to each large star, as Fig. 13.5 shows.
We assume that the stars are spheres as in Fig. 13.2.
We first calculate the magnitudes F_1 and F_2 from Eq. (13.1) and then compute the vector sum by using components:
F_{\mathrm{g}}=\frac{G m_{1} m_{2}}{r^{2}} (13.1)
\begin{aligned}F_{1} &=\frac{\left[\begin{array}{c}\left(6.67 \times 10^{-11} \mathrm{~N}\cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right) \\\times\left(8.00 \times 10^{30}\mathrm{~kg}\right)\left(1.00 \times 10^{30} \mathrm{~kg}\right)\end{array}\right]}{\left(2.00 \times 10^{12} \mathrm{~m}\right)^{2}+\left(2.00 \times 10^{12} \mathrm{~m}\right)^{2}} \\&=6.67 \times 10^{25} \mathrm{~N} \\F_{2} &=\frac{\left[\begin{array}{c}\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right) \\\times\left(8.00 \times 10^{30} \mathrm{~kg}\right)\left(1.00 \times 10^{30} \mathrm{~kg}\right)\end{array}\right]}{\left(2.00 \times 10^{12} \mathrm{~m}\right)^{2}} \\&=1.33\times 10^{26} \mathrm{~N}\end{aligned}The x- and y-components of these forces are
\begin{aligned}&F_{1 x}=\left(6.67 \times 10^{25} \mathrm{~N}\right)\left(\cos 45^{\circ}\right)=4.72 \times 10^{25} \mathrm{~N} \\&F_{1 y}=\left(6.67 \times 10^{25} \mathrm{~N}\right)\left(\sin 45^{\circ}\right)=4.72 \times 10^{25} \mathrm{~N} \\&F_{2 x}=1.33 \times10^{26} \mathrm{~N} \\&F_{2 y}=0\end{aligned}The components of the total force \vec{w} on the small star are
\begin{aligned}&F_{x}=F_{1 x}+F_{2 x}=1.81 \times 10^{26} \mathrm{~N} \\&F_{y}=F_{1 y}+F_{2 y}=4.72 \times 10^{25} \mathrm{~N}\end{aligned}The magnitude of \vec{w} and its angle θ (see Fig. 13.5) are
\begin{aligned}F &=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{\left(1.81 \times 10^{26} \mathrm{~N}\right)^{2}+\left(4.72 \times 10^{25} \mathrm{~N}\right)^{2}} \\&=1.87 \times 10^{26}\mathrm{~N} \\\theta &=\arctan \frac{F_{y}}{F_{x}}=\arctan \frac{4.72 \times 10^{25}\mathrm{~N}}{1.81 \times 10^{26} \mathrm{~N}}=14.6^{\circ}\end{aligned}
EVALUATE: While the force magnitude F is tremendous, the magnitude of the resulting acceleration is not: a = F/m = \left(1.87 \times 10^{26} \mathrm{~N}\right) /\left(1.00 \times 10^{30} \mathrm{~kg}\right)=1.87 \times 10^{-4} \mathrm{~m} / \mathrm{s}^{2}. Furthermore, the force \vec{w} is not directed toward the center of mass of the two large stars.
