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Space Physics - Earth and the Solar System

Grade 11IGCSEPhysics

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

🔑Concepts

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The Solar System consists of the Sun, eight planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune), dwarf planets, asteroids, and comets.

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The four inner planets (Mercury to Mars) are rocky and small, while the four outer planets (Jupiter to Neptune) are gas or ice giants and much larger.

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The Sun is a medium-sized star consisting mostly of hydrogen (HH) and helium (HeHe). It radiates energy due to nuclear fusion of hydrogen into helium in its core.

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Gravitational field strength (gg) varies on different planets. The weight of an object is calculated using W=mgW = mg.

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Orbital motion is maintained by gravitational force acting as a centripetal force. For an object in a circular orbit, the orbital speed vv is constant, but the velocity is constantly changing because the direction of motion is changing.

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Stars form from a giant cloud of dust and gas called a nebula. Gravity collapses the nebula into a protostar, which eventually becomes a main sequence star when nuclear fusion begins.

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The life cycle of a star depends on its mass. Low-mass stars become Red Giants and then White Dwarfs. High-mass stars become Red Supergiants, then explode as a Supernova, leaving behind a neutron star or a black hole.

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Redshift is the observed increase in the wavelength of light from distant galaxies. The further away a galaxy is, the greater the redshift, indicating that the universe is expanding.

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Hubble’s Law states that the recessional velocity vv of a galaxy is proportional to its distance dd from Earth: v=H0dv = H_0 d.

📐Formulae

v=2πrTv = \frac{2 \pi r}{T}

W=mgW = mg

v=H0dv = H_0 d

t≈1H0t \approx \frac{1}{H_0}

💡Examples

Problem 1:

Earth orbits the Sun at an average distance of 1.5×108 km1.5 \times 10^{8} \text{ km}. Given that it takes 365.25365.25 days to complete one orbit, calculate the orbital speed of Earth in m/s\text{m/s}.

Solution:

First, convert distance to meters: r=1.5×1011 mr = 1.5 \times 10^{11} \text{ m}. Convert time to seconds: T=365.25×24×60×60≈3.156×107 sT = 365.25 \times 24 \times 60 \times 60 \approx 3.156 \times 10^{7} \text{ s}. Apply the formula: v=2π(1.5×1011 m)3.156×107 sv = \frac{2 \pi (1.5 \times 10^{11} \text{ m})}{3.156 \times 10^{7} \text{ s}} v≈29865 m/s≈3.0×104 m/sv \approx 29865 \text{ m/s} \approx 3.0 \times 10^{4} \text{ m/s}

Explanation:

Orbital speed is the circumference of the orbit divided by the orbital period. Units must be converted to SI (meters and seconds) for the final answer in m/s\text{m/s}.

Problem 2:

A distant galaxy is moving away from Earth at a velocity of 5.0×106 m/s5.0 \times 10^{6} \text{ m/s}. If the Hubble constant H0H_0 is 2.2×10−18 s−12.2 \times 10^{-18} \text{ s}^{-1}, calculate the distance to the galaxy in meters.

Solution:

Use Hubble's Law: v=H0dv = H_0 d. Rearrange for dd: d=vH0d = \frac{v}{H_0} d=5.0×106 m/s2.2×10−18 s−1d = \frac{5.0 \times 10^{6} \text{ m/s}}{2.2 \times 10^{-18} \text{ s}^{-1}} d≈2.27×1024 md \approx 2.27 \times 10^{24} \text{ m}

Explanation:

Hubble's Law relates the recessional velocity of a galaxy to its distance from the observer. Using the provided constant, we can estimate distances on a cosmological scale.

Earth and the Solar System - Revision Notes & Key Formulas | IGCSE Grade 11 Physics