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.

TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$11.2~\mathrm{km~s^{-1}}$
›Reveal solutionSolution
Escape speed depends on mass and radius. Given constant density, radius scales as the cube root of mass, so escape speed scales as $M^{1/3}$ . With mass 8 times Earth’s, escape speed doubles to $22.4\ \mathrm{km\,s^{-1}}$ .

The escape speed from a planet’s surface is the minimum speed needed for an object to leave its gravitational pull forever. The formula is $v_e = \sqrt{\frac{2GM}{R}}$ , where $M$ is the planet’s mass and $R$ its radius. The key insight: if the mean density $\rho$ is the same as Earth’s, then mass and radius are linked through $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ , so $R \propto M^{1/3}$ . That lets us find how $v_e$ changes when only $M$ changes, without needing numerical values.

Write the escape speed formula.

For Earth: $v_e = \sqrt{\frac{2GM}{R}}$ . For the other planet: $v_e' = \sqrt{\frac{2G M'}{R'}}$ , where $M' = 8M$ and $\rho' = \rho$ .

Relate radius to mass using constant density.

Since $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ is the same for both,

$\frac{M}{R^3} = \frac{M'}{R'^3} \quad\Rightarrow\quad \frac{R'}{R} = \left(\frac{M'}{M}\right)^{1/3} = 8^{1/3} = 2.$
So the planet’s radius is twice Earth’s.

Substitute into the escape speed ratio.

$\frac{v_e'}{v_e} = \sqrt{\frac{M'}{M} \cdot \frac{R}{R'}} = \sqrt{8 \cdot \frac{1}{2}} = \sqrt{4} = 2.$
Therefore $v_e' = 2 \times 11.2\ \mathrm{km\,s^{-1}} = 22.4\ \mathrm{km\,s^{-1}}$ . …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$\dfrac{1}{10}$
›Reveal solutionSolution
The escape velocity depends on both mass and radius. Equating the given ratios leads to the planet’s radius being $\frac{R}{5}$ .

The escape velocity from a celestial body is the minimum speed needed for an object to break free from its gravitational pull without further propulsion. The formula is derived from energy conservation: kinetic energy at launch equals the work done against gravity to infinity. For a planet of mass $M$ and radius $R$ , the escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$ , where $G$ is the universal gravitational constant.

The key insight: escape velocity scales as the square root of mass over radius. So if we know how mass and escape velocity compare between two planets, we can solve for the unknown radius ratio.

Let’s denote Earth’s mass as $M_e$ and radius as $R_e = R$ . The planet’s mass is $M_p = \frac{1}{10} M_e$ . Its escape velocity $v_{e,p}$ is given as $\frac{1}{\sqrt{2}}$ times Earth’s escape velocity $v_{e,e}$ .

Write the escape velocity for Earth:

$v_{e,e} = \sqrt{\frac{2GM_e}{R}}$

Write the escape velocity for the planet:

$v_{e,p} = \sqrt{\frac{2G M_p}{R_p}} = \sqrt{\frac{2G \cdot \frac{1}{10} M_e}{R_p}}$

The problem states:

$v_{e,p} = \frac{1}{\sqrt{2}} v_{e,e}$
Substitute the expressions:
$\sqrt{\frac{2G \cdot \frac{1}{10} M_e}{R_p}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM_e}{R}}$

Square both sides to remove square roots: $\frac{2G \cdot \frac{1}{10} M_e}{R_p} = \frac{1}{2} \cdot \frac{2GM_e}{R}$ …
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TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$x$
›Reveal solutionSolution
The key idea is conservation of angular momentum: if Earth’s radius becomes $x$ times larger, its moment of inertia increases by $x^2$ , so its rotation slows by $x^2$ , making the new period $24x^2$ hours.

We start with the concept: Earth’s rotation is determined by its angular momentum, which is conserved if no external torque acts. When the radius changes, the mass distribution changes, altering the moment of inertia. Since angular momentum $L = I \omega$ stays constant, the angular speed $\omega$ must adjust inversely to $I$ . The period $T = 2\pi / \omega$ then changes proportionally to $I$ .

Moment of inertia of a sphere For a solid sphere of mass $M$ and radius $R$ , the moment of inertia about its axis is $I = \frac{2}{5} M R^2$ . If the radius becomes $x$ times the present value, the new radius is $R' = xR$ , so the new moment of inertia is

$I' = \frac{2}{5} M (xR)^2 = \frac{2}{5} M R^2 \cdot x^2 = I \cdot x^2.$

Conservation of angular momentum No external torque acts (we assume the change happens internally or gradually), so

$L = I \omega = I' \omega'.$
Substituting $I' = I x^2$ gives
$I \omega = (I x^2) \omega' \quad \Rightarrow \quad \omega' = \frac{\omega}{x^2}.$

Relation between period and angular speed The period $T = 2\pi / \omega$ . The new period is
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$Q. Which of the following statement is false? (A) All planets move in elliptical orbits with sun at one of the foci. (B) The square root of time period of revolution of planet is proportional to the cube of semi major axis of the ellipse. (C) Line that joins any planet to the sun sweeps equal areas in equal intervals of time. (D) The measurement of G has refined by Cavendish’s experiment.$
›Reveal solutionSolution
Kepler’s three laws describe planetary motion; statement (B) misstates the third law (it says “square root” instead of “square”), making it the false statement.

The question tests your knowledge of Kepler’s laws of planetary motion and a famous experimental result. Kepler’s laws are:

Law of Ellipses – planets move in ellipses with the Sun at one focus.

Law of Equal Areas – a line joining a planet to the Sun sweeps out equal areas in equal times.

Law of Harmonies – the square of the orbital period is proportional to the cube of the semi-major axis.

Statement (D) refers to Cavendish’s experiment, which measured the gravitational constant $G$ . That is a true historical fact. So the false statement must be among (A)–(C). Let’s check each carefully.

Statement (A): “All planets move in elliptical orbits with sun at one of the foci.”

This is exactly Kepler’s first law. It is true.

Pitfall: Some might think orbits are perfectly circular, but Kepler showed they are ellipses (circles are a special case of ellipses, but the Sun is at a focus, not the center).

Statement (B): “The square root of time period of revolution of planet is proportional to the cube of semi major axis of the ellipse.”

Kepler’s third law says: $T^2 \propto a^3$ , where $T$ is the period and $a$ is the semi-major axis.

Taking square roots: $\sqrt{T^2} = T \propto a^{3/2}$ . That is not “square root of time period proportional to cube of semi-major axis.” The statement says $\sqrt{T} \propto a^3$ , which is false. …
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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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