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Earth, Moon, and the Sun - Changing view of night sky from the Earth

Grade 7CBSE

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

🔑Concepts

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The Earth rotates on its axis from West to East once every 2424 hours, which causes the Sun, Moon, and stars to appear as if they are moving from East to West across the sky.

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The Pole Star (PolarisPolaris) appears stationary in the night sky because it is situated directly above the Earth's axis of rotation in the North direction.

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A constellation is a group of stars that forms a recognizable pattern in the night sky, such as Ursa Major (Great Bear), Orion (The Hunter), and Cassiopeia.

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The Moon revolves around the Earth in an elliptical orbit, taking approximately 27.327.3 days for one revolution, which is known as a sidereal month.

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Phases of the Moon occur because we see only that part of the Moon which reflects sunlight towards us; the cycle from one New Moon to the next takes approximately 29.529.5 days.

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Distances in the universe are measured in Light Years. One light year is the distance light travels in one year at a speed of 3×108 m/s3 \times 10^{8} \text{ m/s}.

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Stars appear to shift their positions in the sky from month to month due to the Earth's revolution around the Sun (36514365 \frac{1}{4} days).

📐Formulae

1 Light Year≈9.46×1012 km1 \text{ Light Year} \approx 9.46 \times 10^{12} \text{ km}

Speed of Light (c)=3×108 m/s\text{Speed of Light } (c) = 3 \times 10^{8} \text{ m/s}

Time Period of Moon’s Phases≈29.5 days\text{Time Period of Moon's Phases} \approx 29.5 \text{ days}

Angle of Earth’s rotation per hour=360∘24=15∘/hour\text{Angle of Earth's rotation per hour} = \frac{360^{\circ}}{24} = 15^{\circ}/\text{hour}

💡Examples

Problem 1:

If a star is located 88 light years away from the Earth, calculate its distance in kilometers using scientific notation.

Solution:

8×9.46×1012 km=75.68×1012 km=7.568×1013 km8 \times 9.46 \times 10^{12} \text{ km} = 75.68 \times 10^{12} \text{ km} = 7.568 \times 10^{13} \text{ km}

Explanation:

We multiply the number of light years by the value of one light year (9.46×1012 km9.46 \times 10^{12} \text{ km}) to find the total distance.

Problem 2:

Why does the Moon rise about 5050 minutes later every day?

Solution:

Daily delay≈360∘29.5 days×24×60 mins360∘≈48.8 minutes\text{Daily delay} \approx \frac{360^{\circ}}{29.5 \text{ days}} \times \frac{24 \times 60 \text{ mins}}{360^{\circ}} \approx 48.8 \text{ minutes}

Explanation:

As the Earth rotates, the Moon also moves forward in its orbit. To bring the Moon back to the same position in the sky, the Earth has to rotate a bit more, which takes approximately 5050 minutes.

Problem 3:

Calculate the total number of days in 44 lunar months (synodic months).

Solution:

29.5×4118.0\begin{array}{r} 29.5 \\ \times 4 \\ \hline 118.0 \end{array}

Explanation:

One synodic month (from New Moon to New Moon) is 29.529.5 days. Multiplying by 44 gives 118118 days.

Changing view of night sky from the Earth Class 7 Notes & Examples