In the figure below, two point particles are fixed on an x axis separated by a distance d. Particle A has mass mA and particle B has mass 4.00mA. A third particle C, of mass 96.0mA, 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 find the position of particle C.
Let's assume that the origin of the x-axis is at the position of particle A, and particle B is located at the position d.
The gravitational force between two point masses can be calculated using Newton's law of gravitation:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force
G is the gravitational constant (approximately 6.674 x 10^-11 Nm^2/kg^2)
m1 and m2 are the masses of the two particles
r is the distance between the two particles
In this case, we want the net gravitational force on particle A from particles B and C to be zero. So, the gravitational force between particle A and particle B must be equal in magnitude and opposite in direction to the gravitational force between particle A and particle C.
The distances between particle A and B, and particle A and C are d and x, respectively.
Now, let's set up the equation for the net gravitational force as follows:
F_AB = F_AC
(G * (mA * 4.00mA)) / d^2 = (G * (mA * 96.0mA)) / x^2
By simplifying the equation, we can solve for x:
4.00 / d^2 = 96.0 / x^2
Cross-multiplying gives:
4.00 * x^2 = 96.0 * d^2
Dividing both sides by 4.00 gives:
x^2 = 24.0 * d^2
Finally, taking the square root of both sides gives:
x = √(24.0 * d^2)
So, the x coordinate at which the net gravitational force on particle A from particles B and C is zero is x = √(24.0 * d^2).