This MCQ module is based on: Satellite Orbits
TOPIC 32 OF 33
Satellite Orbits
🎓 Class 11
Physics
CBSE
Theory
Ch 7 – Gravitation
⏱ ~14 min
🌐 Language: [gtranslate]
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Satellite Orbits
7.11 Earth Satellites
A satellite is any body that revolves around a larger body under its gravitational pull. Earth has one natural satellite — the Moon — and thousands of artificial ones from communication, weather, GPS, military and scientific missions.
For a satellite to remain in a stable circular orbit at altitude h above Earth, the gravitational pull must provide exactly the centripetal force needed:
\[\frac{G M_E m}{(R_E+h)^2} = \frac{m\,v_o^2}{R_E+h}\]
7.12 Orbital Velocity
Solving the equation above gives the orbital velocity:
Orbital Velocity at altitude h:
\[v_o = \sqrt{\frac{G M_E}{R_E+h}}\]
For an orbit very close to Earth (h ≈ 0): \(v_o = \sqrt{gR_E}\approx 7.92\,\text{km/s}\).
Note the elegant relation between orbital and escape velocity at the surface:
\[v_e = \sqrt{2}\,v_o\]
So escape velocity is exactly √2 ≈ 1.414 times the orbital velocity — meaning if a satellite in low orbit (7.9 km/s) is given just 41.4% extra speed (to 11.2 km/s), it escapes Earth's gravity entirely.
7.12.1 Time Period of Satellite
The time for one complete orbit (period T) is circumference divided by speed:
\[T = \frac{2\pi(R_E+h)}{v_o} = 2\pi\sqrt{\frac{(R_E+h)^3}{GM_E}}\]
This is Kepler's Third Law applied to Earth satellites! Squaring: \(T^2\propto (R_E+h)^3\).
7.13 Energy of an Orbiting Satellite
A satellite in a circular orbit has both kinetic and potential energy:
\[KE = \frac{1}{2}m v_o^2 = \frac{1}{2}\frac{G M_E m}{r}\quad,\quad U = -\frac{G M_E m}{r}\]
Where \(r = R_E + h\). The total mechanical energy is:
Total Energy of orbiting satellite:
\[E = KE + U = \frac{1}{2}\frac{GM_E m}{r}-\frac{GM_E m}{r}=-\frac{1}{2}\frac{GM_E m}{r}\]
\[\boxed{E = -\frac{1}{2}\frac{GM_E m}{r}}\]
The total energy is negative (system is bound) and equals exactly half the (negative) PE. Also: \(KE = -E\), and \(U = 2E\).
To "boost" a satellite to a higher orbit, you must increase its total energy (less negative). Counter-intuitively, the speed in the higher orbit is smaller, but the PE has risen even more, so the total E is greater (less negative).
7.14 Geostationary and Polar Satellites
7.14.1 Geostationary Satellite
A geostationary satellite orbits Earth in the equatorial plane with period T = 24 hours, matching Earth's rotation. From the ground it appears motionless — perfect for TV broadcasting and telecommunications. Setting T = 86,400 s and solving Kepler's law:
\[r = \left(\frac{GM_E T^2}{4\pi^2}\right)^{1/3} \approx 4.22\times10^7\,\text{m}\]
Subtracting Earth's radius gives altitude h ≈ 35,786 km. India's INSAT, GSAT and educational satellite EDUSAT all operate in geostationary orbit.
7.14.2 Polar Satellite
A polar satellite is in a low (~500–800 km) near-polar orbit. Each orbit takes ~100 min, and because Earth rotates underneath, the satellite scans different longitudes on successive passes — covering the entire globe in ~24 hours. India's IRS (Indian Remote Sensing) series uses polar orbits for cartography, agriculture, ocean monitoring and disaster management.
7.15 Weightlessness
The weight of a body is defined as the contact force (e.g., the reading of a weighing machine) needed to support it. When a person stands on a weighing scale on Earth, the scale pushes up with N = mg, and reads weight = mg.
In a freely-falling lift, the floor falls with the same g as the person — no contact force is needed, so the scale reads zero. The same happens in orbit: the satellite (and everything inside) falls toward Earth, but tangential motion keeps "missing" the planet, producing a perpetual free fall. Hence astronauts float, pens float, water forms spheres — all in apparent weightlessness despite gravity still being strong (≈8.7 m/s² at ISS altitude).
Interactive Simulation: Orbital Velocity Calculator
Slide the orbit altitude to compute orbital velocity and period.
v_orbital = 7.67 km/s | T = 92.7 min
Try h = 35,786 km — you'll see T ≈ 1436 min (= 23.93 h) for geostationary orbit.
Worked Example 1: ISS orbital speed
The ISS orbits at h = 400 km. Find its orbital velocity and period. (M_E = 6×10²⁴ kg, R_E = 6400 km)
r = 6400 + 400 = 6800 km = 6.8×10⁶ m
\[v_o = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{6.8\times 10^6}}\]
\[ = \sqrt{5.88\times 10^7} \approx \boxed{7670\,\text{m/s} = 7.67\,\text{km/s}}\]
\[T = \frac{2\pi r}{v_o} = \frac{2\pi\times 6.8\times 10^6}{7670} \approx 5571\,\text{s}\approx \boxed{93\,\text{min}}\]
The ISS makes about 15.5 orbits each day.
Worked Example 2: Energy needed to lift an ISS satellite
How much extra energy must be supplied to a 1000-kg satellite to raise it from h = 400 km to h = 800 km?
Total energy at radius r: \(E = -\frac{GMm}{2r}\)
\[\Delta E = E_2 - E_1 = -\frac{GMm}{2}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]
With r1 = 6800 km, r2 = 7200 km:
\[\Delta E = -\frac{6.67\times10^{-11}\times 6\times 10^{24}\times 1000}{2}\left(\frac{1}{7.2\times 10^6}-\frac{1}{6.8\times 10^6}\right)\]
\[= -2.00\times10^{17}\times(-8.17\times 10^{-9}) \approx \boxed{1.63\times 10^9\,\text{J}}\]
1.63 GJ — equivalent to about 36 kg of jet fuel.
Activity 7.4 — Newton's Cannon Thought ExperimentL4 Analyse
Imagine a cannon on a very tall mountain firing horizontally with adjustable muzzle speed.
- Low speed: cannonball follows a parabolic path and hits the ground close by.
- Higher speed: it travels farther before landing.
- At exactly orbital speed (≈7.9 km/s near surface, ignoring air drag): the curvature of its path matches the curvature of Earth — it "falls around" the Earth and never lands.
- Beyond orbital speed: elliptical orbit.
- At escape speed (11.2 km/s): the path becomes a parabola, never returning.
Predict: At what fraction of escape speed does an object just orbit?
Orbital speed = v_e / √2 ≈ 0.707 × v_e. So once you've spent the energy to reach 70.7% of escape speed, just one more push of about 41.4% gets you all the way out of Earth's gravitational hold.
Competency-Based Questions
India's GSAT-30 communication satellite operates from a geostationary orbit at ~35,786 km altitude, providing telecommunications across India. Meanwhile, the Cartosat-3 polar satellite captures high-resolution Earth-imaging data from ~509 km altitude.
Q1. The orbital velocity of a satellite is inversely proportional to:L2 Understand
Answer: (a) √r. v_o = √(GM/r), so v_o ∝ 1/√r. Larger r ⇒ smaller v_o.
Q2. Geostationary satellites are placed only above the equator. Why?L4 Analyse
A geostationary orbit must lie in the equatorial plane so that the satellite rotates around Earth's spin axis with the same period as Earth's rotation. Any tilted orbit would cause the sub-satellite point to oscillate north–south, so it would not appear stationary.
Q3. The total mechanical energy of an orbiting satellite is negative. State the physical meaning.L2 Understand
A negative total energy means the satellite is bound to the central body. To liberate it (set E = 0 at infinity) we must supply extra energy equal to |E|. If E ≥ 0, the orbit becomes a parabola or hyperbola and the body escapes.
Q4. Fill in the blank: A polar satellite covers the whole Earth in ____ hours because Earth rotates under its orbit.L1 Remember
~24 hours. A polar satellite's orbit is nearly fixed in inertial space; Earth rotates 360° in 24 h, so successive orbits map adjacent longitude strips, eventually scanning the whole surface.
Q5. HOT: Design a satellite-network architecture (number, orbit type, spacing) to provide continuous Internet to every point on Earth.L6 Create
Sample design (Starlink-like): Use ~4000+ small satellites in low Earth orbit (~550 km, T ≈ 95 min) arranged in 72 polar/quasi-polar orbital planes with 22 satellites per plane. Each satellite has ~ 1000 km ground coverage; with planes 5° apart in longitude, every point on Earth is in view of at least 3 satellites simultaneously. Inter-satellite laser links create a mesh. Trade-offs: LEO requires many satellites (vs few GEO), but each has 50× lower latency (~30 ms vs ~600 ms) and 10× lower path loss — making real-time gaming and video calls possible.
Assertion–Reason Questions
(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
A: Astronauts inside a spacecraft orbiting Earth feel weightless.
R: Gravity is zero at the orbital altitude.
(C). A is TRUE; R is FALSE. Gravity at 400 km is still ~8.7 m/s². Weightlessness arises from continuous free fall, not absence of gravity.
A: A satellite in a higher orbit has greater total energy.
R: The total mechanical energy of a satellite is E = −GMm/(2r), which is less negative for larger r.
(A). Both true; R explains A. "Greater energy" means a less negative number.
A: Geostationary satellites must rotate from west to east.
R: The Earth rotates from west to east, and a geostationary satellite must match Earth's rotation to stay above one point.
(A). Both true; R explains A perfectly.
Frequently Asked Questions - Satellite Orbits
What is the main concept covered in Satellite Orbits?
In NCERT Class 11 Physics Chapter 7 (Gravitation), "Satellite Orbits" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Satellite Orbits useful in real-life applications?
Real-life applications of Satellite Orbits from NCERT Class 11 Physics Chapter 7 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Satellite Orbits?
Key formulas in Satellite Orbits (NCERT Class 11 Physics Chapter 7 Gravitation) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 7?
NCERT Class 11 Physics Chapter 7 (Gravitation) is structured so each part builds on the previous one. Satellite Orbits connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Satellite Orbits?
CBSE board questions from Satellite Orbits typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Satellite Orbits lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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