krit.club logo

Space Physics - Earth and the Solar System

Grade 12IGCSEPhysics

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

🔑Concepts

The Earth rotates on its axis once every 2424 hours, which causes the cycle of day and night as different parts of the planet face the Sun.

The Earth orbits the Sun once every 365.25365.25 days. The Earth's axis is tilted at an angle of approximately 23.523.5^\circ, which results in the changing seasons as the Earth moves around its orbit.

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

The four inner planets (Mercury to Mars) are rocky and metallic (terrestrial), while the four outer planets (Jupiter to Neptune) are larger and composed mainly of gas and ice (gas giants).

Gravitational field strength (gg) varies across different planets and depends on the mass (MM) and radius (rr) of the planet; for example, gEarth9.8 m/s2g_{Earth} \approx 9.8\text{ m/s}^2 while gMoon1.6 m/s2g_{Moon} \approx 1.6\text{ m/s}^2.

The Sun's energy is released through nuclear fusion in its core, where Hydrogen nuclei fuse to form Helium: 4 11H 24He+2 +10e+energy4\ ^1_1H \rightarrow \ ^4_2He + 2\ ^0_{+1}e + \text{energy}.

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}. This unit is used to measure interstellar and intergalactic distances.

The orbital speed of a planet or moon is determined by the radius of its circular orbit (rr) and its orbital period (TT).

📐Formulae

v=2πrTv = \frac{2\pi r}{T}

W=mgW = mg

1 light-year (ly)9.46×1015 m1\text{ light-year (ly)} \approx 9.46 \times 10^{15}\text{ m}

gsurface=GMr2g_{surface} = \frac{GM}{r^2}

💡Examples

Problem 1:

Calculate the average orbital speed of the Earth around the Sun. Assume the orbit is circular with a radius of 1.5×1011 m1.5 \times 10^{11}\text{ m} and an orbital period of 365.25365.25 days.

Solution:

First, convert the period TT into seconds: T=365.25×24×60×603.156×107 sT = 365.25 \times 24 \times 60 \times 60 \approx 3.156 \times 10^7\text{ s}. Next, use the formula v=2πrTv = \frac{2\pi r}{T}. Substituting the values: v=2×π×1.5×10113.156×1072.99×104 m/sv = \frac{2 \times \pi \times 1.5 \times 10^{11}}{3.156 \times 10^7} \approx 2.99 \times 10^4\text{ m/s}.

Explanation:

The orbital speed is the total distance of the circular path (2πr2\pi r) divided by the time taken for one full revolution (TT).

Problem 2:

An astronaut has a mass of 75 kg75\text{ kg} on Earth. Calculate their weight on Mars, where the gravitational field strength is gMars=3.7 N/kgg_{Mars} = 3.7\text{ N/kg}.

Solution:

Use the formula W=mgW = mg. Given m=75 kgm = 75\text{ kg} and g=3.7 N/kgg = 3.7\text{ N/kg}, the weight is: W=75×3.7=277.5 NW = 75 \times 3.7 = 277.5\text{ N}.

Explanation:

Mass is a constant property of the object, while weight depends on the local gravitational field strength of the celestial body.

Earth and the Solar System - Revision Notes & Key Formulas | IGCSE Grade 12 Physics