A small mass M and a small mass 3M are 1.00 m apart. Where should you put a third small mass so that the net gravitational force on it due to the other two masses is zero?
To determine where to place a third small mass so that the net gravitational force on it due to the other two masses is zero, we can use the concept of gravitational attraction.
The gravitational force between two objects is given by Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
where F represents the gravitational force, G is the gravitational constant (6.67430 x 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.
In this scenario, we have a small mass M and a small mass 3M placed 1.00 m apart. We need to find the position for a third small mass where the net gravitational force on it equals zero.
Let's assume that the small mass M is located at the origin (0, 0) on a coordinate plane. Thus, the small mass 3M is located at coordinates (1.00 m, 0).
Now, we can introduce the coordinates of the third small mass as (x, y). To find where to place the third mass, we need to determine the x and y coordinates that satisfy the condition of zero net gravitational force.
The net gravitational force acting on the third small mass is the vector sum of the gravitational forces due to the other two masses. In magnitude, it can be expressed as:
|F_net| = |F1| + |F2|
Since we want the net gravitational force to be zero:
|F_net| = |F1| + |F2| = 0
Applying Newton's law of universal gravitation to the first small mass M:
F1 = G * (M * m3) / R1^2
where m3 is the mass of the third small mass and R1 is the distance between the first small mass and the third small mass.
Applying Newton's law of universal gravitation to the second small mass 3M:
F2 = G * ((3M) * m3) / R2^2
where R2 is the distance between the second small mass and the third small mass.
Since the two forces have opposite directions, we can set them equal and consider their magnitudes:
|F1| = |F2|
G * (M * m3) / R1^2 = G * ((3M) * m3) / R2^2
Canceling out the gravitational constant G and the mass m3 from both sides:
(M / R1^2) = ((3M) / R2^2)
Simplifying the equation leads to:
1 / R1^2 = 3 / R2^2
Taking the square root of both sides, we get:
R2 / R1 = sqrt(3)
Since R1 is 1.00 m, we can solve for R2:
R2 = R1 * sqrt(3)
= 1.00 m * sqrt(3)
≈ 1.73 m
So, the third small mass should be placed at coordinates (1.73 m, 0) to achieve zero net gravitational force.
Gravitational force
=GMm/r²
So for the mass m between masses M and 3M, one metre apart, equate the gravitational forces:
GMm/x² = G(3M)m/(1-x)²
Solve for x.
[(√3-1)/2]
do it this way
M(x)=3M(100-x)
x=300-3x
4x=300
x=75cm
or 0.75m