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.

CBSE 2025Set ANNUAL1 markMCQ
$Q. The correct relation between the time period of a planet (T) and its mean distance (R) from the Sun is$
(a) (A) T² ∝ R
(b) (B) T³ ∝ R²
(c) (C) T² ∝ R³
(d) (D) T ∝ R³
›Reveal solutionSolution
[!TLDR]

(C) T² ∝ R³

Why

Kepler's third law of planetary mot …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The orbital radius of any satellite of the earth is four times the orbital radius of a geo-stationary satellite. Then the time period of revolution of that satellite of the earth will be equal to (A) 8 days (B) 4 days (C) 16 days (D) 24 hours$
›Reveal solutionSolution

Orbital radius 4× geostationary radius ⇒ period = 8× geostationary period = 8 days.

Kepler's third law: $T^2\propto r^3$ , so $T\propto r^{3/2}$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. According to Kepler's law of period (T) :$
(a) T α R²
(b) T α R³
(c) T α R¹ᐟ²
(d) T α R³ᐟ²
›Reveal solutionSolution

Kepler's third law: the square of the orbital period is proportional to the cube of the semi-major axis, so T ∝ R³ᐟ².

Kepler's law of periods (third law) states that for planets orbiting the sun (or, more generally, any satellite in a near-circular orbit):

$T^2 \propto R^3$

Taking the square root of both sides:

$T \propto R^{3/2}$

…

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CBSE 2024Set ANNUAL1 mark
$Q. State Kepler's law of periods.$
›Reveal solutionSolution

Kepler's third law relates the orbital period of a planet to the size of its orbit: $T^2 \propto a^3$ .

Kepler's law of periods (Kepler's third law) states that the square of the time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of its elliptical orbit:

$T^2 \propto a^3$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The period of moon's rotation around the earth is nearly 29 days. If moon's mass were 2 fold its present value, and all other things remain unchanged, the period of Moon's rotation would be nearly$
(1) 29 sqrt(2) days
(2) 29 / sqrt(2) days
(3) 29 x 2 days
(4) 29 days
›Reveal solutionSolution

In the two-body approximation where the central body's mass (Earth) dominates, the orbiting body's own mass drops out of the period formula entirely - so changing the Moon's mass doesn't change its orbital period.

For a body of mass m orbiting a much more massive central body of mass M (here, Moon orbiting Earth), equating gravitational force to the required centripetal force gives:

GMm/r^2 = m (4 pi^2 / T^2) r

Notice the orbiting body's mass m appears on BOTH sides and cancels out completely:

GM/r^2 = 4 pi^2 r / T^2

T^2 = 4 pi^2 r^3 / (GM)

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Kepler's third law states that the rotation period of planet around the sun is:$
(a) T^2 ∝ r^3
(b) T ∝ r^3
(c) T ∝ r^(3/4)
(d) T ∝ r
›Reveal solutionSolution

Kepler's third law: T^2 ∝ r^3.

Kepler's law of periods states that the square of the orbital period T of a planet is directly proportional to the cube of the semi-major axis (mean orbital radi …

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CBSE 2020Set ANNUAL1 markMCQ
$Q. Kepler's third law is also known as:$
(a) Law of orbits
(b) Law of areas
(c) Law of periods
(d) None of these
›Reveal solutionSolution

Kepler's three laws each have a name: the first is the law of orbits, the second is the law of areas, and the third (T^2 proportional to a^3) is the law of periods.

Kepler's first law (law of orbits): planets move in elliptical orbits with the sun at one focus.

Kepler's second law (law of areas): the radius vector from sun to planet sweeps equal areas in equal time intervals. …

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CBSE 2019Set ANNUAL1 mark
$Q. What is the mathematical form of Kepler's third law?$
›Reveal solutionSolution

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

Kepler's third law, also called the Law of Periods, states: the square of the time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.

Mathematically:

$T^2 \propto r^3 \quad \Rightarrow \quad \frac{T^2}{r^3} = \text{constant (same for all planets orbiting the Sun)}$

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The time period of an earth satellite in circular orbit is independent of$
(a) the mass of the satellite
(b) radius of its orbit.
(c) both the mass of satellite and radius of the orbit.
(d) neither the mass of satellite nor the radius of the orbit.
›Reveal solutionSolution

Setting gravitational force equal to centripetal force for an orbiting satellite, the satellite's mass cancels out algebraically, so the orbital time period depends only on orbital radius and the mass of the central body (planet), never on the satellite's own mass.

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

GMm/r^2 = m v^2/r = m (4 pi^2 r / T^2)

The satellite's mass m appears on both sides and cancels:

GM/r^2 = 4 pi^2 r / T^2 …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Kepler third law related to the planetary motion is:$
(a) T ∝ r
(b) T ∝ r²
(c) T ∝ r³
(d) T ∝ r^(3/2)
›Reveal solutionSolution

Kepler's third law: T² ∝ r³, which on taking a square root becomes T ∝ r^{3/2}.

Kepler's third law of planetary motion states that the square of the time period T of a planet's revolution around the Sun is directly proportional to the cube of the semi-major axis (mean orbital radius) r of its orbit:

$T^2 \propto r^3$

Taking the square root of both sides:

$T \propto r^{3/2}$

…

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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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