What is the relation between orbital velocity and escape velocity?

For a satellite orbiting close to Earth's surface (r ≈ R), orbital speed is v_o = √(g R) ≈ 7.92 km/s. The escape speed from Earth's surface is v_e = √(2 g R) ≈ 11.2 km/s. Therefore, v_e = √2 v_o ≈ 1.414 v_o. This proves that if the orbital speed of a circulating satellite is increased by (√2 - 1) * 100% = 41.4%, the satellite will escape Earth's gravitational pull and leave its orbit.

Why do astronauts experience weightlessness in an orbiting satellite?

Astronauts experience weightlessness not because gravity is zero in orbit (gravity at 400 km altitude is still about 89% of surface gravity), but because the satellite and the astronaut are both in continuous free fall towards the center of the Earth with the exact same gravitational acceleration g_h. Since both accelerate together, the normal contact reaction force between the astronaut and the floor/scale is zero (N = 0).

§ 7.4NCERT Class 11 · Physics · Chapter 7

Satellite Motion, Orbital Velocity, Energy & Satellite Types

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 8 min read

A satellite is any natural or artificial celestial body revolving around a planet in a stable gravitational orbit. Understanding satellite kinematics, energy relations, and the physical distinction between geostationary and polar orbits forms an essential pillar of classical mechanics and aerospace physics.

1. Orbital Velocity (vo) and Period of Revolution

Consider a satellite of mass m orbiting Earth (mass M, radius R) at an altitude h in a circular orbit of radius r = R + h. The gravitational attraction provides the necessary centripetal force:

m v_o² / r = G M m / r²

v_o = √(G M / r) = √[G M / (R + h)] = R √[g / (R + h)]

Orbital Speed Near Earth's Surface (h ≪ R)

When the satellite orbits very close to Earth's surface (h ≈ 0, r ≈ R):

v_o = √(g R) = √(9.8 × 6.4 × 10⁶) ≈ 7.92 km/s ≈ 8 km/s

Relation Between Orbital Velocity and Escape Speed

v_e = √(2 g R) = √2 v_o ≈ 1.414 v_o
If the speed of a circular satellite is increased by 41.4% (Δv / v_o = √2 − 1 ≈ 0.414), its total mechanical energy becomes zero and it escapes into parabolic orbit!

Period of Revolution (T)

The time taken by the satellite to complete one full revolution is:

T = 2πr / v_o = 2π (R + h) / √[G M / (R + h)] = 2π √[(R + h)³ / (G M)]

Squaring both sides confirms Kepler's Third Law: T² = [4π² / (G M)] r³ ∝ r³.

For a near-Earth orbit (h ≪ R): T = 2π √(R / g) ≈ 84.6 minutes ≈ 1.4 hours.

Satellite Orbital Energy & Orbit Classification

Interrelationship between kinetic, potential, and total mechanical energy, plus orbital classifications

ORBITAL ENERGY RELATIONSHIPS:
-----------------------------
Kinetic Energy (K)        = + G M m / (2r)   [Always Positive]
Potential Energy (U)      = - G M m / r      [Always Negative, Double magnitude of K]
Total Energy (E)          = - G M m / (2r)   [Always Negative for bound orbit]

Key Identity: E = -K = U / 2  <-->  Binding Energy (B.E.) = -E = + G M m / (2r)

GEOSTATIONARY vs POLAR ORBITS:
------------------------------
GEOSTATIONARY ORBIT (Communication):            POLAR ORBIT (Remote Sensing):
- Orbit: Equatorial plane, West to East        - Orbit: Over North and South poles
- Altitude h: ~35,800 km (~36,000 km)          - Altitude h: ~500 to 800 km
- Period T: Exactly 24 hours                   - Period T: ~100 minutes
- Stationary relative to Earth observers       - Scans strips of rotating Earth beneath

Energy Conservation Identity: E = −K = U / 2. If a satellite encounters atmospheric drag and loses total energy (ΔE < 0), r decreases, potential energy becomes more negative, but its kinetic energy and speed v_o actually increase!

2. Energy of an Orbiting Satellite & Binding Energy

Energy Type	Mathematical Formulation	Sign & Relative Magnitude
Kinetic Energy (K)	K = ½ m v_o² = G M m / [2(R + h)]	Strictly positive (+K)
Potential Energy (U)	U = −G M m / (R + h)	Negative (−2K)
Total Mechanical Energy (E)	E = K + U = −G M m / [2(R + h)]	Negative (−K = U/2)
Binding Energy (B.E.)	B.E. = −E = +G M m / [2(R + h)]	Positive work needed to free satellite

Trajectory Classification by Total Mechanical Energy (E):

E < 0: Bound orbit (Circular if v = v_o; Elliptical if v_o < v < v_e).
E = 0: Escapes along a Parabolic trajectory (v = v_e).
E > 0: Escapes along a Hyperbolic trajectory (v > v_e).

3. Geostationary vs Polar Satellites

1. Geostationary (Geosynchronous / Parking) Satellites

Time Period: Exactly equal to Earth's rotational period, T = 24 hours.
Orbital Plane: Must lie strictly in the equatorial plane of the Earth.
Direction of Revolution: West to East (same direction as Earth's rotation).
Orbital Altitude (h): Setting T = 24 h in Kepler's formula yields r = (G M T² / 4π²)^1/3 ≈ 42,200 km. Subtracting R = 6,400 km yields: h ≈ 35,800 km ≈ 36,000 km.
Applications: Telecommunication, television broadcast, and global weather surveillance.

2. Polar Satellites

Orbital Plane: Revolves in a North-South orbit passing over the Earth's geographic poles.
Orbital Altitude (h): Low Earth Orbit (LEO) at h ≈ 500 – 800 km.
Time Period: T ≈ 100 minutes (crosses any given latitude multiple times daily).
Applications: High-resolution environmental remote sensing, meteorology, military surveillance, and cartography.

4. Weightlessness in Satellites

A passenger inside an orbiting space vehicle feels completely weightless. The apparent weight of an object is measured by the normal reaction force N exerted on it by the supporting surface.

For an astronaut of mass m standing on a scale inside an orbiting satellite at height h:

F_g − N = m a_c ⇒ m g_h − N = m (v_o² / r)

Since the centripetal acceleration is supplied entirely by gravity (v_o² / r = g_h):

m g_h − N = m g_h ⇒ N = 0

Because the astronaut and the spacecraft accelerate together towards Earth's center with acceleration g_h, the normal contact reaction vanishes. This condition of continuous free-fall is termed weightlessness.

5. Solved High-Yield JEE / NEET Practice Questions

Question 1: If the kinetic energy of a circular orbiting satellite is E_k, then its total mechanical energy is:

(A) + E_k

(B) − E_k

(D) −2 E_k

Show answer

Correct Answer: (B) − E_k

Step-by-step Solution:
1. In a circular orbit of radius r, K = G M m / (2r) = E_k.
2. Potential energy is U = −G M m / r = −2 E_k.
3. Total mechanical energy is E = K + U = E_k − 2 E_k = − E_k.
Hence, total energy is always equal in magnitude to kinetic energy, but with a negative sign.

Question 2: An artificial satellite is revolving in an orbit of radius r with speed v. If its speed is increased by 41.4%, the satellite will:

(A) Move to an orbit of radius 2r

(B) Escape from Earth's gravitational field

(D) Continue in the same orbit

Show answer

Correct Answer: (B) Escape from Earth's gravitational field

Step-by-step Solution:
1. An increase of 41.4% means the new speed is v' = v + 0.414 v = 1.414 v = √2 v.
2. The escape speed at any radius r is v_e = √2 v_o.
3. When speed reaches √2 v_o, the total mechanical energy E becomes zero:
E' = ½ m (√2 v_o)² − G M m / r = m v_o² − G M m / r = 0.
4. With zero total energy, the satellite escapes from Earth's gravitational field along a parabolic trajectory.

Question 3: The height of a geostationary satellite above the Earth's surface is approximately:

(A) 6,400 km

(B) 18,000 km

(D) 42,000 km

Show answer

Correct Answer: (C) 36,000 km

Step-by-step Solution:
1. From T² = [4π² / (G M)] (R + h)³ with T = 24 h = 86,400 s:
R + h ≈ 42,200 km (orbital radius from center).
2. The altitude above Earth's surface is h = 42,200 km − 6,400 km ≈ 35,800 km ≈ 36,000 km.

6. Frequently Asked Questions (FAQs)

1. Can a geostationary satellite be placed in an orbit passing over New Delhi or London?

No! A geostationary satellite must orbit strictly in the equatorial plane of the Earth. A satellite orbiting over a specific non-equatorial latitude would have its orbital center at Earth's center of mass, so its plane would intersect both northern and southern hemispheres, meaning it would oscillate north and south rather than staying stationary above one point.

2. What happens to the period of revolution of a satellite when its orbital radius increases?

By Kepler's Third Law (T² ∝ r³), increasing the orbital radius r increases the period of revolution T and decreases the orbital speed (v_o ∝ 1/√r).

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