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Earth and Space - The Solar System and Planets

Grade 6IB

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

🔑Concepts

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The Solar System is a gravitationally bound system consisting of the Sun and the objects that orbit it. The Sun contains approximately 99.86%99.86\% of the system's total mass.

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The eight planets are divided into two groups: Inner planets (Mercury, Venus, Earth, Mars) which are terrestrial and rocky, and Outer planets (Jupiter, Saturn, Uranus, Neptune) which are gas giants or ice giants composed largely of H2H_2, HeHe, and CH4CH_4.

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Gravity (FgF_g) is the fundamental force that keeps planets in elliptical orbits around the Sun. The strength of this force is determined by the masses (mm) of the objects and the distance (rr) between them.

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The Astronomical Unit (AUAU) is a simplified unit of distance where 1 AU1 \text{ AU} represents the average distance from the center of the Earth to the center of the Sun, approximately 1.5×108 km1.5 \times 10^8 \text{ km}.

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Planetary motion involves two main types: Rotation (spinning on an axis, defining a 'day') and Revolution (orbiting the Sun, defining a 'year').

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The Atmosphere of planets varies: Venus has a thick atmosphere of CO2CO_2 causing a runaway greenhouse effect, while Mars has a thin atmosphere mostly of CO2CO_2 with traces of O2O_2 and ArAr.

📐Formulae

1 AU≈1.5×108 km1 \text{ AU} \approx 1.5 \times 10^8 \text{ km}

Fg=Gm1m2r2F_g = G \frac{m_1 m_2}{r^2}

Orbital Speed (v)=2πrT\text{Orbital Speed } (v) = \frac{2\pi r}{T}

Weight=m×g\text{Weight} = m \times g

💡Examples

Problem 1:

Light travels at a speed of approximately 3×105 km/s3 \times 10^5 \text{ km/s}. If the distance from the Sun to Earth is 1.5×108 km1.5 \times 10^8 \text{ km}, calculate how many minutes it takes for sunlight to reach Earth.

Solution:

t=dv=1.5×108 km3×105 km/s=500 secondst = \frac{d}{v} = \frac{1.5 \times 10^8 \text{ km}}{3 \times 10^5 \text{ km/s}} = 500 \text{ seconds} 50060≈8.33 minutes\frac{500}{60} \approx 8.33 \text{ minutes}

Explanation:

By using the formula for time t=dvt = \frac{d}{v}, we divide the total distance by the speed of light. Converting the result from seconds to minutes gives the standard '8-minute' light travel time.

Problem 2:

The planet Jupiter is located roughly 5.2 AU5.2 \text{ AU} from the Sun. Express this distance in kilometers using scientific notation.

Solution:

d=5.2×(1.5×108 km)=7.8×108 kmd = 5.2 \times (1.5 \times 10^8 \text{ km}) = 7.8 \times 10^8 \text{ km}

Explanation:

To convert Astronomical Units to kilometers, multiply the given AUAU value by the value of 1 AU1 \text{ AU} (1.5×108 km1.5 \times 10^8 \text{ km}).

Problem 3:

If an astronaut has a mass of 70 kg70 \text{ kg} and the acceleration due to gravity on Mars is gMars≈3.7 m/s2g_{Mars} \approx 3.7 \text{ m/s}^2, what is the astronaut's weight on Mars?

Solution:

W=m×g=70 kg×3.7 m/s2=259 NW = m \times g = 70 \text{ kg} \times 3.7 \text{ m/s}^2 = 259 \text{ N}

Explanation:

Mass remains constant regardless of location, but weight changes based on the local gravitational acceleration (gg). Using W=mgW = mg, we find the force in Newtons (NN).

The Solar System and Planets - Revision Notes & Key Formulas | IB Grade 6 Science