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Physics - Space and the Solar System

Grade 8IGCSE

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, their moons, and smaller bodies like asteroids, comets, and dwarf planets. The planets in order from the Sun are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.

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Gravity is the force of attraction between all masses. The strength of the gravitational field (gg) determines the weight of an object on a planet's surface. On Earth, g≈9.8 N/kgg \approx 9.8 \text{ N/kg} (often rounded to 10 N/kg10 \text{ N/kg} in IGCSE Grade 8).

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Orbital motion is maintained by gravitational force. For a circular orbit, the orbital speed (vv) is calculated using the circumference of the orbit (2πr2\pi r) divided by the orbital period (TT).

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Mass is the amount of matter in an object and is measured in kg\text{kg}, while weight is a force measured in Newtons (N\text{N}). Weight changes depending on the gravitational field strength: W=mgW = mg.

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Distances in space are vast. A light-year is the distance light travels in one year, approximately 9.46×1015 m9.46 \times 10^{15} \text{ m}. Within the solar system, we often use the Astronomical Unit (AU\text{AU}), where 1 AU≈1.5×108 km1 \text{ AU} \approx 1.5 \times 10^{8} \text{ km}.

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The Sun is a medium-sized star that produces energy through nuclear fusion, primarily converting Hydrogen (1H^{1}H) into Helium (4He^{4}He).

📐Formulae

W=m×gW = m \times g

v=2πrTv = \frac{2 \pi r}{T}

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

1 ly=c×(365.25×24×60×60) where c≈3×108 m/s1 \text{ ly} = c \times (365.25 \times 24 \times 60 \times 60) \text{ where } c \approx 3 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

Calculate the weight of a robot with a mass of 50 kg50 \text{ kg} on the surface of Mars, where the gravitational field strength is gMars=3.7 N/kgg_{Mars} = 3.7 \text{ N/kg}.

Solution:

W=m×g=50 kg×3.7 N/kg=185 NW = m \times g = 50 \text{ kg} \times 3.7 \text{ N/kg} = 185 \text{ N}

Explanation:

Weight is the product of mass and the local gravitational field strength. While the mass remains 50 kg50 \text{ kg} everywhere, the weight decreases on Mars compared to Earth because gg is lower.

Problem 2:

A satellite orbits a planet at a distance of 8000 km8000 \text{ km} from the planet's center. If it takes 2 hours2 \text{ hours} to complete one full orbit, calculate its orbital speed in km/h\text{km/h}.

Solution:

v=2πrT=2×3.14159×8000 km2 h≈25132.7 km/hv = \frac{2 \pi r}{T} = \frac{2 \times 3.14159 \times 8000 \text{ km}}{2 \text{ h}} \approx 25132.7 \text{ km/h}

Explanation:

The distance traveled in one orbit is the circumference of the circle (2πr2 \pi r). Dividing this distance by the time period (TT) gives the constant orbital speed.

Problem 3:

If a star is located 4.24.2 light-years away from Earth, calculate this distance in meters.

Solution:

D=4.2×(9.46×1015 m)≈3.97×1016 mD = 4.2 \times (9.46 \times 10^{15} \text{ m}) \approx 3.97 \times 10^{16} \text{ m}

Explanation:

One light-year is the distance light travels in a vacuum in one year. To find the total distance, multiply the number of light-years by the value of one light-year in meters.