A satellite is to be placed in equatorial geostationary orbit around earth for communication. (a) Calculate he

NCERT Exemplar · Q36

Q. A satellite is to be placed in equatorial geostationary orbit around earth for communication.

(a) Calculate height of such a satellite.

(b) Find out the minimum number of satellites that are needed to cover entire earth so that at least one satellites is visible from any point on the equator. [

M = 6 \times 10^{24}

kg,

R = 6400

km,

T = 24

G = 6.67 \times 10^{-11}

SI units]

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Concept understanding — Kepler's Third Law

Kepler's Third Law: The Harmony of the Planets

Imagine you're watching two planets orbiting the Sun. One is close in — Mercury, zipping around in just 88 days. Another is far out — Saturn, taking nearly 30 years to complete one lap. You'd expect the farther planet to take longer, but here's the surprising part: the relationship isn't just "farther = slower." It's much more precise, and it reveals a deep truth about gravity itself.

The Intuition

Think of a planet as a runner on a circular track. The farther out the track, the longer the lap — that's obvious. But Kepler noticed something subtler: if you double the distance from the Sun, the orbital period doesn't just double. It increases by a factor of about $2.8$ (which is $\sqrt{8}$ ). Triple the distance, and the period grows by about $5.2$ (which is $\sqrt{27}$ ).

There's a pattern here. The period seems to grow as the 3/2 power of the distance. Why? Because gravity weakens with distance, so a farther planet feels a weaker pull and moves more slowly — not just because the track is longer, but because it's moving slower along that track.

The Precise Statement

T^2 \propto a^3

For planets orbiting the Sun, if you measure $T$ in Earth years and $a$ in astronomical units (AU, where 1 AU = Earth's average distance from the Sun), the constant of proportionality is exactly 1:

T^2 = a^3

So for Earth: $T = 1$ year, $a = 1$ AU, and $1^2 = 1^3$ — it checks out.

For Mars: $a \approx 1.52$ AU, so $T^2 = (1.52)^3 \approx 3.51$ , giving $T \approx 1.87$ years. That's about 687 days — exactly right.

Note

This law applies to any body orbiting a much more massive central body: moons around planets, satellites around Earth, binary stars around each other. The constant of proportionality changes depending on the mass of the central body.

Why It Works (The Physics)

Newton later showed that Kepler's Third Law is a direct consequence of his law of gravitation. For a circular orbit (a good approximation for most planets), the centripetal force needed to keep the planet in orbit is provided by gravity:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Here $M$ is the Sun's mass, $m$ the planet's mass, $r$ the orbital radius, and $v$ the orbital speed. The speed is related to the period by $v = 2\pi r / T$ . Substituting and simplifying:

\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}

Rearranging:

T^2 = \frac{4\pi^2}{GM} r^3

The quantity $4\pi^2/(GM)$ is a constant for all planets orbiting the Sun. So $T^2 \propto r^3$ — exactly Kepler's law.

Important

The constant $4\pi^2/(GM)$ depends only on the mass of the central body. This means: if you know the period and distance of any moon or planet, you can calculate the mass of the body it orbits. This is how astronomers "weigh" stars, black holes, and galaxies.

A Common Mistake …

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Matching gravity to the centripetal requirement for a 24-hour orbit gives an orbital radius of about 42,300 km, i.e. a height of about 35,900 km above Earth's surface. Since a satellite is visible only from within a limiting angle set by $\cos\theta=R/r$ , one satellite covers about $163°$ of the equator, so a minimum of 3 geostationary satellites, spaced $120°$ apart, are needed to keep every equatorial point in view of at least one.

Part (a): Height of the geostationary orbit

For a satellite of mass $m$ in a circular orbit of radius $r$ , gravity supplies the centripetal force:

\frac{GMm}{r^2} = m\omega^2r = m\left(\frac{2\pi}{T}\right)^2r

Solving for $r$ :

r^3 = \frac{GMT^2}{4\pi^2}

With $G=6.67\times10^{-11}$ , $M=6\times10^{24}$ kg, and $T=24\text{ h}=86400$ s:

r^3 = \frac{(6.67\times10^{-11})(6\times10^{24})(86400)^2}{4\pi^2} \approx 7.57\times10^{22}\text{ m}^3

r \approx 4.23\times10^7\text{ m} = 42{,}300\text{ km}

Height above the surface:

h = r-R = 42{,}300-6{,}400 = 35{,}900\text{ km}

Part (b): Minimum number of satellites

A point on the equator can see a satellite only if the satellite is above its horizon. At the extreme visible point $P$ , the line of sight $PS$ (to the satellite $S$ ) is tangent to the Earth, so the triangle formed with Earth's centre $C$ has a right angle at $P$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$Q. State Kepler's third law of planetary motion.$
›Reveal solutionSolution

Kepler's third law (the law of periods) states that the square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit.

Statement: The square of the time period (T) of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis (a) of its elliptical orbit.

Mathematically: T² ∝ a³, or T²/a³ = constant (the same constant for all planets orbiting the Sun).

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$Q. Question number 33 (continued) is a case-study based question. Read the case carefully and answer the question that follows: Recently, on 7th September 2025, a lunar eclipse was observed on the day of full moon. Some people in a village avoided eating or cooking during the eclipse, believing it could bring ill effects. A young boy, who had studied about eclipses in science, knew that solar and lunar eclipses are natural celestial phenomena caused by the motion of the Earth, Moon, and Sun and have no harmful effects on living beings. When he explained this to his mother, she ignored his reasoning and followed the traditional belief. On his way to a friend's house, he noticed people performing prayers and rituals to ward off the so-called "ill effects" of the eclipse. The boy felt disheartened at the superstitions and decided to study more about the scientific reason behind such phenomena.$
(iii) The time period of Jupiter around the Sun is 10 years. Calculate its distance from the Sun, given that the Earth's distance from the Sun is R and its time period is 1 year. OR Alternative case study (Or), continued: Ajay was always fascinated by the flights of airplanes and rockets. He often wondered how rockets could travel into space and not return, while everything he threw upward would always fall back due to Earth's gravitational pull. Curious about this, he discussed the matter with his elder brother, who explained that rockets are launched based on a scientific principle. When the rocket is projected with increasing speed, a point comes when its speed becomes high enough to overcome Earth's gravitational pull — this speed is called the escape velocity. At this speed, the rocket can move out of Earth's gravitational field and never return on its own. Understanding this concept inspired Ajay to dream of becoming an astronaut in the future.
(iii) Find the ratio of escape velocity to orbital velocity for a satellite revolving close to the earth's surface.
›Reveal solutionSolution

Jupiter's distance from the Sun is about 4.64R, where R is the Earth's distance from the Sun.

(This is the primary part of the case-study question; the alternative — the ratio of escape velocity to orbital velocity — is not required since the primary is fully answerable.)

By Kepler's third law, T² ∝ r³ for planets orbiting the Sun, where T is the orbital period and r is the (mean) orbital radius.

For Earth: T_E = 1 year, r_E = R.

For Jupiter: T_J = 10 years, r_J = ?

Taking the ratio:

(T_J / T_E)² = (r_J / R)³

…

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Q.A satellite is to be placed in equatorial geostationary orbit around earth for communication.

Part (a): Height of the geostationary orbit

Part (b): Minimum number of satellites

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