Question

A white dwarf and a black hole are in a binary system. The white dwarf has the same mass as the Sun, while the black hole is 15 times more massive than the Sun. The two are separated by a distance of 4 AU. How much time do they take to orbit the centre of mass?

a) 1 year
b) 2 years
c) 3 years
d) 4 years

Answer 1

The correct orbital period for the white dwarf and the black hole in a binary system with a 4 AU separation is approximately 2 years. (b)

Step-by-step explanation:

In a binary system, the orbital period can be calculated using Kepler's third law, which relates the orbital period (P) to the semi-major axis (a) of the orbit. For this case, the masses of the two objects are crucial. Given that the white dwarf has the same mass as the Sun and the black hole is 15 times more massive than the Sun, the combined mass is 16 times the mass of the Sun.

The formula for the orbital period in a binary system is
`\(P^2 = \frac{{4π^2}}{{G(M_1 + M_2)}}a^3\)`, where P is the orbital period, G is the gravitational constant, M1 and M2 are the masses of the two objects, and a is the semi-major axis.

In this scenario, the combined mass (M1 + M2) is 16 times the mass of the Sun, and the distance between them (a) is given as 4 astronomical units (AU). Plugging these values into the formula, we get
`\(P^2 = \frac{{4π^2}}{{G(16M_(\odot))}}(4 \text{ AU})^3\)`.

Calculating the orbital period gives us
`\(P^2 = \frac{{64π^2}}{{G(16M_(\odot))}}(64 \text{ AU}^3)\)`, and when simplified,
`\(P^2 = \frac{{4π^2}}{{G(M_(\odot))}}(4 \text{ AU})^3\)`.

Therefore,
`\(P^2 = \frac{{1}}{{M_(\odot)}} * 4π^2 * 4^3 = 16π^2\)` years squared, which simplifies to
`\(P = 4√(π^2)\)` years, which is approximately equal to 2 years.(b)