A binary star consists of two dense sphereaty masses of 10rest power 30 kg and 2 x 10rest power 30kg. Whose centres are 10rest power7km apart and which rotate togethe with angula velocity w about an axis which intersects a line joining their centres - show that if the only forces acting on the stars are gravitational the axis of rotation passes through the center of man system - find the value of w.
To show that the axis of rotation passes through the center of mass system, we need to verify that the torques due to gravitational forces acting on each mass cancel each other out.
Let's start by considering the torque due to the gravitational force acting on the first mass, M1. The gravitational force between the two masses is given by Newton's Law of Gravitation:
F = G * (M1 * M2) / r^2
Where:
F is the gravitational force between the masses
G is the universal gravitational constant
M1 and M2 are the masses of the stars
r is the distance between the centers of the stars
The torque, τ1, due to this force acting on M1 can be calculated as:
τ1 = r * F * sin(θ)
Where:
θ is the angle between the line connecting the centers of the stars and the axis of rotation.
The torque due to the gravitational force acting on mass M2, τ2, can be calculated in a similar manner.
Since the masses are rotating together with an angular velocity, ω, the torque due to gravitational forces should cancel each other out to keep the system in rotational equilibrium:
τ1 + τ2 = 0
Using the values and equations given, we can rewrite this equation in terms of the known values:
r * F * sin(θ1) + r * F * sin(θ2) = 0
Simplifying further:
r * F * (sin(θ1) + sin(θ2)) = 0
Since the forces act on opposite sides of the axis, the angles θ1 and θ2 are equal in magnitude but opposite in direction:
θ1 = -θ2
Therefore, the equation becomes:
r * F * (sin(θ1) - sin(θ1)) = 0
This equation is satisfied if sin(θ1) - sin(θ1) = 0, which is always true. Hence, the torques due to gravitational forces cancel each other out, and the axis of rotation passes through the center of mass system.
To find the value of ω (angular velocity), we can use the principle of conservation of angular momentum. In the absence of external torques, the total angular momentum of the system remains constant:
L = I * ω
Where:
L is the angular momentum
I is the moment of inertia
For a binary system consisting of two spheres rotating about an axis passing through the center of mass system, the moment of inertia I can be calculated as:
I = (1/5) * (M1 * r^2 + M2 * r^2)
Substituting the given values:
I = (1/5) * [(10^30 kg) * (10^7 km)^2 + (2 * 10^30 kg) * (10^7 km)^2]
Converting km to meters:
I = (1/5) * [(10^30 kg) * (10^7 * 10^3 m)^2 + (2 * 10^30 kg) * (10^7 * 10^3 m)^2]
Calculating the moment of inertia:
I = (1/5) * [(10^30 kg) * (10^10 m)^2 + (2 * 10^30 kg) * (10^10 m)^2]
I = (1/5) * [10^20 kg * (10^10 m)^2 + 2 * 10^20 kg * (10^10 m)^2]
I = (1/5) * [10^20 kg * 10^20 m^2 + 2 * 10^20 kg * 10^20 m^2]
I = (1/5) * [10^40 kg * m^2 + 2 * 10^40 kg * m^2]
I = (1/5) * [3 * 10^40 kg * m^2]
I = 3/5 * 10^40 kg * m^2
Finally, we can solve for ω using the conservation of angular momentum equation:
(10^30 kg) * (10^7 km) * w = (3/5) * 10^40 kg * m^2 * ω
Converting km to meters:
(10^30 kg) * (10^7 * 10^3 m) * ω = (3/5) * 10^40 kg * m^2 * ω
Simplifying further:
(10^30 kg) * (10^10 m) * ω = (3/5) * 10^40 kg * m^2 * ω
Cancelling out the terms:
(10^10) * ω = (3/5) * 10^10 * m
Simplifying further:
ω = (3/5) m
Therefore, the value of ω (angular velocity) is (3/5) times the value of m.