krit.club logo

Physics - The Solar System

Grade 9IGCSE

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

🔑Concepts

•

The Solar System consists of the Sun, eight planets, dwarf planets (like Pluto), moons, asteroids, and comets.

•

The Sun is a medium-sized star composed mostly of Hydrogen and Helium. It produces energy through nuclear fusion: 411H→24He+energy4 {}^1_1H \to {}^4_2He + \text{energy}.

•

Planets orbit the Sun in elliptical paths. The gravitational pull of the Sun provides the centripetal force required to keep planets in orbit.

•

The planets are divided into two groups: Inner (rocky) planets (Mercury, Venus, Earth, Mars) and Outer (gaseous/ice giants) planets (Jupiter, Saturn, Uranus, Neptune).

•

Orbital speed is the speed at which an object orbits around another body, calculated using the orbital circumference and the time period (TT): v=2πrTv = \frac{2 \pi r}{T}.

•

Gravitational field strength (gg) varies between planets depending on their mass and radius. On Earth, g≈9.8 N/kgg \approx 9.8 \text{ N/kg}.

•

A light-year is the distance light travels in a vacuum in one year. Since the speed of light c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}, one light-year is approximately 9.5×1015 m9.5 \times 10^{15} \text{ m}.

•

The further a planet is from the Sun, the lower its orbital speed and the longer its orbital period (TT).

📐Formulae

v=2πrTv = \frac{2 \pi r}{T}

W=mgW = mg

Distance=speed of light×time\text{Distance} = \text{speed of light} \times \text{time}

ρ=mV\rho = \frac{m}{V}

💡Examples

Problem 1:

The Earth orbits the Sun at an average distance of 1.5×1011 m1.5 \times 10^{11} \text{ m}. Calculate the orbital speed of the Earth in m/s\text{m/s}, assuming a circular orbit and a time period of 365.25 days365.25 \text{ days}.

Solution:

  1. Convert time to seconds: T=365.25×24×60×60=31,557,600 sT = 365.25 \times 24 \times 60 \times 60 = 31,557,600 \text{ s}.
  2. Use the formula: v=2πrTv = \frac{2 \pi r}{T}.
  3. Substitute values: v=2×3.142×1.5×101131,557,600v = \frac{2 \times 3.142 \times 1.5 \times 10^{11}}{31,557,600}.
  4. v≈29,865 m/sv \approx 29,865 \text{ m/s}.

Explanation:

To find the speed, we divide the total circumference of the orbit (2πr2 \pi r) by the time taken to complete one full revolution in seconds.

Problem 2:

An astronaut has a mass of 70 kg70 \text{ kg}. If the gravitational field strength on Mars is 3.7 N/kg3.7 \text{ N/kg}, calculate the astronaut's weight on Mars.

Solution:

W=m×gW = m \times g W=70 kg×3.7 N/kgW = 70 \text{ kg} \times 3.7 \text{ N/kg} W=259 NW = 259 \text{ N}

Explanation:

Weight is the force of gravity acting on an object's mass. It is calculated by multiplying the mass (constant everywhere) by the local gravitational field strength.