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Keeping Time with the Skies - Why Do We Launch Artificial Satellites into Space?

Grade 8CBSE

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

🔑Concepts

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Artificial satellites are man-made objects launched from Earth that revolve around it or other celestial bodies. Unlike the Moon (a natural satellite), these are designed for specific tasks like communication and research.

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The first Indian satellite was Aryabhata, launched in 1975. Since then, ISRO (Indian Space Research Organisation) has launched numerous series like INSAT (communication) and IRS (remote sensing).

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Geostationary Satellites: These orbit the Earth at an altitude of approximately 36,000 km36,000\text{ km} above the equator. Their orbital period is T=24 hoursT = 24\text{ hours}, matching Earth's rotation, making them appear stationary in the sky.

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Polar Satellites: These orbit from north to south, passing over the Earth's poles. They are closer to Earth (altitudes of 500 km500\text{ km} to 800 km800\text{ km}) and are used for weather forecasting and remote sensing.

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Satellites stay in orbit due to a balance between their forward velocity (inertia) and the Earth's gravitational pull. The force of gravity provides the necessary centripetal force for circular motion, defined by F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}.

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Applications of satellites include: 1) Radio and TV transmission, 2) Telecommunication, 3) Weather monitoring, 4) Navigation (GPS/NavIC), and 5) Scientific exploration of space.

📐Formulae

F=GMmr2F = G \frac{M m}{r^2}

v=GMrv = \sqrt{\frac{G M}{r}}

T=2πrvT = \frac{2 \pi r}{v}

T2r3=4π2GM\frac{T^2}{r^3} = \frac{4 \pi^2}{G M}

💡Examples

Problem 1:

A satellite is orbiting Earth at a distance rr. If the distance from the center of the Earth is increased to 2r2r, what happens to the gravitational force FF between the Earth and the satellite?

Solution:

The force of gravity follows the inverse square law: F∝1r2F \propto \frac{1}{r^2}. If the new distance r′=2rr' = 2r, then the new force F′F' is F′=GMm(2r)2=GMm4r2=14FF' = G \frac{M m}{(2r)^2} = G \frac{M m}{4r^2} = \frac{1}{4}F.

Explanation:

Because the distance is squared in the denominator, doubling the distance reduces the gravitational pull to one-fourth of its original value.

Problem 2:

Calculate the difference in altitude between a Geostationary satellite (altitude ≈35,786 km\approx 35,786\text{ km}) and a Low Earth Orbit (LEO) satellite (altitude ≈2,000 km\approx 2,000\text{ km}).

Solution:

Difference = 35,786 km−2,000 km35,786\text{ km} - 2,000\text{ km} 35786−200033786\begin{array}{r} 35786 \\ -2000 \\ \hline 33786 \end{array} The difference is 33,786 km33,786\text{ km}.

Explanation:

To find the distance between two orbital heights, we subtract the lower altitude from the higher altitude.

Problem 3:

If a satellite travels a distance of d=2πrd = 2 \pi r in one full orbit, and its orbital period is TT, express its speed vv.

Solution:

Speed is defined as distance divided by time. For one orbit: v=2πrTv = \frac{2 \pi r}{T}

Explanation:

The circumference of the circular path is 2πr2 \pi r. Dividing this by the time TT taken for one revolution gives the average orbital speed.