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 where the net gravitational force on particle A is zero, we need to set up an equation based on the gravitational forces between the three particles.
Let's assume the x coordinate of particle A is xA, the x coordinate of particle B is xB, and the x coordinate of particle C is xC.
The gravitational force between two point particles can be given by:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two particles, and r is the distance between them.
First, let's consider the gravitational force between particle A and B. Since the masses of A and B are mA and 4mA respectively, the force can be expressed as:
FA_B = G * (mA * 4mA) / d^2
The negative sign indicates that this force is attractive and acts in the opposite direction.
Now, let's consider the gravitational force between particle A and C. The mass of particle C is 80mA, and the distance between A and C is given by (d - xC). Therefore, the force can be expressed as:
FA_C = G * (mA * 80mA) / (d - xC)^2
Again, the negative sign indicates that this force is attractive and acts towards A.
Since the net gravitational force on particle A is zero, the sum of the forces FA_B and FA_C must be equal to zero:
FA_B + FA_C = 0
Substituting the respective expressions:
G * (mA * 4mA) / d^2 + G * (mA * 80mA) / (d - xC)^2 = 0
Now, we can solve this equation to find the value of xC:
(mA * 4mA) / d^2 + (mA * 80mA) / (d - xC)^2 = 0
Simplifying the equation and solving for xC will yield the x coordinate where particle C should be placed for the net gravitational force on particle A to be zero.