Splitting The Sun In Two

We would like to calculate the net gravitational force on the Earth in two scenarios:

• When the Earth is a distance of $1 \text{AU}$ from the Sun; this net force is defined as $F_1$ in the problem.

• When the Earth is a distance of $1 \text{AU}$ from two stars, each with half the total mass of the Sun; this net force is defined as $F_2$ in the problem.

The problem asks us to calculate the ratio $\frac{F_2}{F_1}$ . We will have to use Newton's Theory of Gravitation to calculate these forces.

Newton's theory states that the magnitude of the gravitational force applied by one mass $M$ on another mass $m$ separated by a distance $R$ is $F = \frac{G m M}{R^2}$ where $G$ is the the gravitational constant. The direction of this force, which is crucial to solving this problem, is along the line connecting the centers of the two masses. Gravity is an attractive force, so the force on mass $m$ is directed toward the other mass $M$ .

Let's start by expressing $F_1$ in terms of the mass of the Earth $m_\oplus$ , the mass of the sun $M_\odot$ and the separation distance $R = 1 \text{AU}$ . In an Earth-Sun system, there is only one gravitational force acting on the Earth, so the magnitude of the net force is $F_1 = \frac{G m_\oplus M_\odot}{R^2}.$

Now we will calculate $F_2$ , the magnitude of the net gravitational force on Earth when the Sun is split into two stars to form a binary system. The net force $F_2$ requires a bit more analysis because each star applies a gravitational force on $m_\oplus$ , and these two forces are not in the same direction, so the net force is the vector sum of two forces.

Let's call the magnitude of the force applied by one of the stars on Earth $f$ . We observe that each star applies a force of the same magnitude because the two stars have identical masses ( $M_\odot/2$ ) and they are equidistant from Earth. Applying Newton’s gravitational law again, we find $f = \frac{G m_\oplus M_\odot}{2 R^2} = \frac{1}{2} F_1.$ We conclude that the magnitude of the gravitational force applied by one of the stars is exactly half of $F_1$ . However, the two forces applied are not in the same direction; they are directed toward the centers of each star as show in this figure:

To finish the calculation of $F_2$ , we will add these two forces acting on the Earth using vector addition. This requires setting up a coordinate system and breaking each of the two forces into components. The diagram above depicts a suggestion for one choice of coordinate system, with the $x$ -axis parallel to the line connecting the centers of the two stars. Since the Earth and two stars are on the vertices of an equilateral triangle, the force pictured on the left ( $f_a$ ) makes an angle $4\pi/3$ with the positive $x$ -axis. Therefore, its components are $f_{a,x} = \frac{1}{2} F_1 \cos (4\pi/3) = -\frac{1}{4} F_1$ and $f_{a,y} = \frac{1}{2} F_1 \sin (4\pi/3) = -\frac{\sqrt{3}}{4} F_1.$ Likewise, the force pictured on the right ( $f_b$ ) in the diagram makes an angle $5\pi/3$ with the positive $x$ -axis, and therefore, it has components $f_{b,x} = \frac{1}{2} F_1 \cos (5\pi/3) = \frac{1}{4} F_1$ and $f_{b,y} = \frac{1}{2} F_1 \sin (5\pi/3) = -\frac{\sqrt{3}}{4} F_1.$

Adding the $x$ -components of the two forces yields $F_{2,x} = f_{a,x} + f_{b,x} = 0 \text{N}$ since they are equal but opposite. However, the $y$ -components do not cancel: $F_{2,y} = f_{a,y} + f_{b,y} = -2\frac{\sqrt{3}}{4} F_1.$ Therefore the magnitude $F_2 = \sqrt{F_{2,x}^2 + F_{2,y}^2} = \frac{\sqrt{3}}{2} F_1.$ Dividing by $F_1$ yields the ratio we set out to calculate: $\frac{F_2}{F_1} = \frac{\sqrt{3}}{2} \approx 0.87.$

Therefore, we calculated that the net gravitational force felt by Earth when it is equidistant from two stars whose total mass equals the Sun’s mass is about $87\%$ of the force felt by Earth when it is the same distance from the Sun.