For a binary system with stars of masses 5 and \(10 \mathrm{M}_{\odot},\) in circular orbits with a period of \(3 \mathrm{yr},\) what is the total energy of the system?

Kepler's Third Law is a crucial principle in orbital mechanics. It relates the period of an orbit to the size of the orbit itself.
Specifically, it states: 'The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.'
Mathematically, it is expressed as:\[ a^3 = \frac{G (M_1 + M_2) T^2}{4 htfa ptrs^2} \]
Here,

• \(a\) is the semi-major axis,
• \(G\) is the gravitational constant,
• \(M_1\) and \(M_2\) are the masses of the two stars,
• and \(T\) is the orbital period.

By understanding and utilizing Kepler's Third Law, we can calculate the semi-major axis given the masses of the stars and the orbital period, which is essential for determining the total energy of a binary star system.

A binary star system consists of two stars that orbit around their common center of mass. These systems are very common in the universe.
The dynamics of binary systems are governed by the laws of gravity, specifically Newton's law of gravitation.
In a binary system:

• Each star's gravity affects the other,
• their orbits can be circular or elliptical,
• the more massive star will typically be closer to the center of mass.

For example, in our exercise, we have one star with mass \(M_1 = 5 \,\text{M}_\odot\), and another with \(M_2 = 10 \,\text{M}_\odot\).
They orbit each other in a circular path with a period of 3 years. Understanding these orbits allows us to determine the energies involved.

Gravitational potential energy (GPE) in a binary star system arises due to the gravitational attraction between the stars.
The formula for GPE in such a system is given by:\[ U = - \frac{G M_1 M_2}{r} \]
here,

• \(U\) is the gravitational potential energy,
• \(G\) is the gravitational constant,
• \(M_1\) and \(M_2\) are the masses of the two stars,
• and \(r\) is the distance between the centers of the two stars.

For a binary star system in a circular orbit, this distance is the same as the semi-major axis \(a\).
In step 4 of the exercise, we use the semi-major axis calculated using Kepler's Third Law to determine the total gravitational potential energy of the system. This step is crucial as it allows us to proceed to find the total energy.

Orbital mechanics, also known as celestial mechanics, deals with the motions of celestial objects under the influence of gravitational forces.
In the context of a binary star system, this involves calculating parameters like the orbital period, semi-major axis, and the energies associated with the orbit.
The total mechanical energy of a binary star system in a circular orbit is given by:\[ E_{total} = K + U \]
where,

• \(K\) is the kinetic energy of the system,
• and \(U\) is the gravitational potential energy.

In a bound system like a binary star system, the total energy is negative, which indicates that the system is gravitationally bound.
This can be simplified to: \[ E_{total} = - \frac{G M_1 M_2}{2a} \]
This negative sign shows that the system's total energy is less than zero, which confirms it is bound. The calculations done in step 5 of the exercise use this formula to find the total energy of the binary system.