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Physics: Space and Astrophysics - Black Holes and Neutron Stars

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Neutron Star is the collapsed core of a massive star (between 1010 and 2525 solar masses, M⊙M_{\odot}) that has undergone a supernova explosion.

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Neutron stars are composed almost entirely of neutrons and are incredibly dense; a single teaspoon of neutron star material would have a mass of about 10910^9 tonnes. Their density ρ\rho is approximately 1017 kg/m310^{17} \text{ kg/m}^3.

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Pulsars are highly magnetized, rotating neutron stars that emit beams of electromagnetic radiation out of their magnetic poles.

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A Black Hole is formed when the core of an extremely massive star (greater than 25M⊙25 M_{\odot}) collapses to a point of infinite density called a Singularity.

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The Event Horizon is the boundary around a black hole beyond which the escape velocity vev_e exceeds the speed of light c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}.

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The distance from the singularity to the event horizon is known as the Schwarzschild Radius (RsR_s).

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Spaghettification is the vertical stretching and horizontal compression of objects falling into a black hole caused by extreme tidal forces.

📐Formulae

Rs=2GMc2R_s = \frac{2GM}{c^2}

ρ=MV\rho = \frac{M}{V}

V=43πr3V = \frac{4}{3}\pi r^3

c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

Calculate the Schwarzschild radius RsR_s of a black hole with a mass 10 times that of our Sun (M=10×1.989×1030 kgM = 10 \times 1.989 \times 10^{30} \text{ kg}). Use G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2 and c=3×108 m/sc = 3 \times 10^8 \text{ m/s}.

Solution:

Rs=2×(6.67×10−11)×(1.989×1031)(3×108)2R_s = \frac{2 \times (6.67 \times 10^{-11}) \times (1.989 \times 10^{31})}{(3 \times 10^8)^2} Rs=2.653×10219×1016R_s = \frac{2.653 \times 10^{21}}{9 \times 10^{16}} Rs≈29,481 m≈29.5 kmR_s \approx 29,481 \text{ m} \approx 29.5 \text{ km}

Explanation:

By substituting the mass and physical constants into the Schwarzschild formula, we find that a 10-solar-mass black hole has an event horizon radius of approximately 29.5 km29.5 \text{ km}.

Problem 2:

If a neutron star has a mass M=3×1030 kgM = 3 \times 10^{30} \text{ kg} and a radius r=10,000 mr = 10,000 \text{ m}, calculate its average density ρ\rho.

Solution:

First, find the volume VV: V=43π(10,000)3≈4.19×1012 m3V = \frac{4}{3} \pi (10,000)^3 \approx 4.19 \times 10^{12} \text{ m}^3 Now, find the density: ρ=3×10304.19×1012≈7.16×1017 kg/m3\rho = \frac{3 \times 10^{30}}{4.19 \times 10^{12}} \approx 7.16 \times 10^{17} \text{ kg/m}^3

Explanation:

Density is mass divided by volume. Using the volume of a sphere, we see that the neutron star's density is extremely high, consistent with theoretical expectations of 1017 kg/m310^{17} \text{ kg/m}^3.