Question

The point masses m and 2m lies along the x-axis, with m at the origin and 2m at x=L. A third point mass M is moved along the x-axis. (a) At what point is the net gravitational force on M due to the other two masses equal to zero? (b) Sketch the x-component of the net force on M due to m and 2m, taking quantities to the right positive. Include the region x<0, 0L. Be especially careful to show the behavior of the graph on either side of x=0 and x=L.

Answer 1

The point where the net gravitational force on the third mass M is zero is at the midpoint between the two masses m and 2m, which is at x=L/2. The x-component of the net force on M due to m and 2m changes sign as we move from left to right across the interval x=0 to x=L, with a zero value at x=L/2.

Step-by-step explanation:

To find the point where the net gravitational force on the third mass M is zero, we can use the concept that the net force is equal to the vector sum of the forces due to the other two masses. The gravitational force between M and m can be calculated using Newton's universal law of gravitation: F₁ = G*M*m / r₁², where G is the gravitational constant, M and m are the masses, and r₁ is the distance between their centers of mass. Similarly, the gravitational force between M and 2m can be calculated using the same formula as F₂ = G*M*(2m) / r₂². At the point where the net force is zero, F₁ and F₂ will have equal magnitudes but opposite directions. From the symmetry of the problem, we can deduce that this occurs at the midpoint between m and 2m, at x=L/2.

To sketch the x-component of the net force on M due to m and 2m, we can first consider the individual forces due to each mass separately. The force due to m is attractive and points towards the origin, while the force due to 2m is repulsive and points away from x=L. As we move closer to m, the magnitude of the force due to m increases, while the magnitude of the force due to 2m decreases. At the midpoint between m and 2m, the magnitudes of both forces are equal. As we move further away from m towards 2m, the magnitude of the force due to m decreases, while the magnitude of the force due to 2m increases. Therefore, the x-component of the net force on M will change sign as we move from left to right across the interval x=0 to x=L, with a zero value at x=L/2.

Answer 2

The net gravitational force on point mass M due to the other two masses will be zero at a point on the x-axis where the gravitational forces from both masses cancel each other out. We can determine this point by setting up an equation using Newton's law of gravitation. To sketch the x-component of the net force on M due to m and 2m, we need to consider the direction of the gravitational force from each mass and find the point where the forces cancel each other out.

Step-by-step explanation:
(a) The net gravitational force on point mass M due to the other two masses will be zero at a point on the x-axis where the gravitational forces from both masses cancel each other out. Let's assume this point is at x = d. Using Newton's law of gravitation, we can set up the equation:
GMm/(d^2) - G(2M)m/((L - d)^2) = 0
Simplifying, we get:

d^3 - Ld^2 + 2Ld - L^2 = 0
Solving this equation will give us the value of d, which represents the point where the net gravitational force is zero.
(b) To sketch the x-component of the net force on M due to m and 2m, we need to consider the direction of the gravitational force from each mass. The force from m will always be attractive and positive, while the force from 2m will be attractive for x < L and repulsive for x > L. The net force will be zero at the point where the forces cancel each other out, which we determined in part (a) to be at x = d.