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Space Physics - Stars and the Universe

Grade 11IGCSEPhysics

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

🔑Concepts

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The Sun is a medium-sized star that releases energy through the nuclear fusion of hydrogen (11H^1_1H) into helium (24He^4_2He).

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A light-year is the distance light travels in a vacuum in one year, approximately 9.5×1015 m9.5 \times 10^{15} \text{ m}.

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Stars are formed from interstellar clouds of dust and gas (nebulae) containing mostly hydrogen.

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The life cycle of a star like the Sun: Protostar →\rightarrow Main Sequence →\rightarrow Red Giant →\rightarrow Planetary Nebula →\rightarrow White Dwarf.

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The life cycle of a massive star: Protostar →\rightarrow Main Sequence →\rightarrow Red Supergiant →\rightarrow Supernova →\rightarrow Neutron Star or Black Hole.

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Redshift is the observed increase in the wavelength of light from distant galaxies, indicating they are moving away from us (v∝Δλv \propto \Delta \lambda).

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Hubble’s Law states that the recession speed vv of a galaxy is directly proportional to its distance dd from Earth.

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The Big Bang Theory is supported by Redshift (expanding universe) and the existence of Cosmic Microwave Background Radiation (CMBR).

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The Hubble constant (H0H_0) represents the ratio of the recession speed of a galaxy to its distance from us, currently estimated at H0≈2.2×10−18 s−1H_0 \approx 2.2 \times 10^{-18} \text{ s}^{-1}.

📐Formulae

v=2πrTv = \frac{2 \pi r}{T}

v=H0dv = H_0 d

t≈1H0t \approx \frac{1}{H_0}

Speed of light (c)≈3.0×108 m/s\text{Speed of light } (c) \approx 3.0 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

Calculate the orbital speed of the Earth around the Sun, assuming a circular orbit with a radius of 1.5×1011 m1.5 \times 10^{11} \text{ m} and an orbital period of 3.15×107 s3.15 \times 10^7 \text{ s}.

Solution:

v=2π(1.5×1011 m)3.15×107 s≈2.99×104 m/sv = \frac{2 \pi (1.5 \times 10^{11} \text{ m})}{3.15 \times 10^7 \text{ s}} \approx 2.99 \times 10^4 \text{ m/s}

Explanation:

The orbital speed vv is calculated by dividing the circumference of the circular orbit 2πr2\pi r by the time period TT for one full revolution.

Problem 2:

A distant galaxy is moving away from Earth at a speed of 4.4×106 m/s4.4 \times 10^6 \text{ m/s}. Given H0=2.2×10−18 s−1H_0 = 2.2 \times 10^{-18} \text{ s}^{-1}, calculate the distance to this galaxy.

Solution:

d=vH0=4.4×106 m/s2.2×10−18 s−1=2.0×1024 md = \frac{v}{H_0} = \frac{4.4 \times 10^6 \text{ m/s}}{2.2 \times 10^{-18} \text{ s}^{-1}} = 2.0 \times 10^{24} \text{ m}

Explanation:

According to Hubble's Law, the distance dd is the recession velocity vv divided by the Hubble constant H0H_0.

Problem 3:

Estimate the age of the universe in years using the Hubble constant H0=2.2×10−18 s−1H_0 = 2.2 \times 10^{-18} \text{ s}^{-1}.

Solution:

t=1H0=12.2×10−18 s−1≈4.55×1017 st = \frac{1}{H_0} = \frac{1}{2.2 \times 10^{-18} \text{ s}^{-1}} \approx 4.55 \times 10^{17} \text{ s}. Converting to years: 4.55×10173.15×107≈1.44×1010 years\frac{4.55 \times 10^{17}}{3.15 \times 10^7} \approx 1.44 \times 10^{10} \text{ years}

Explanation:

The age of the universe can be estimated as the reciprocal of the Hubble constant, which represents the time since all matter was at a single point.

Stars and the Universe - Revision Notes & Key Formulas | IGCSE Grade 11 Physics