One dimension. In Fig. 13-32, two point particles are fixed on an x axis separated by distance d. Particle A has mass mA and particle B has mass 4.00 mA. A third particle C, of mass 80.0 mA, is to be placed on the x axis and near particles A and B. In terms of distance d, at what x coordinate should C be placed so that the net gravitational force on particle A from particles B and C is zero?
To find the x-coordinate at which the net gravitational force on particle A from particles B and C is zero, we need to consider the gravitational forces between them.
The gravitational force between two masses is given by Newton's Law of Universal Gravitation:
F = G * (m1 * m2) / r^2
Where F is the gravitational force, G is the gravitational constant (approximately 6.67430 x 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the two particles, and r is the distance between them.
For particle A and particle B, the gravitational force is directed towards each other and its magnitude is:
F_AB = G * (mA * (4.00mA)) / d^2
For particle A and particle C, the gravitational force is also directed towards each other and its magnitude is:
F_AC = G * (mA * (80.0mA)) / (x^2)
To find the x-coordinate at which the net gravitational force on particle A is zero, the magnitudes of F_AB and F_AC should be equal but opposite in direction:
F_AB = - F_AC
Substituting the expressions for F_AB and F_AC and solving for x, we get:
G * (mA * (4.00mA)) / d^2 = G * (mA * (80.0mA)) / (x^2)
Simplifying the equation, we have:
4.00mA / d^2 = 80.0mA / (x^2)
Cross-multiplying, we get:
4.00mA * (x^2) = 80.0mA * (d^2)
Dividing both sides by 4.00mA, we get:
x^2 = 20.0 * (d^2)
Taking the square root of both sides, we obtain:
x = sqrt(20.0) * d
Therefore, the x-coordinate at which the net gravitational force on particle A from particles B and C is zero is given by x = sqrt(20.0) * d.