The Orbital Velocity Formula gives the minimum speed a satellite needs to maintain a stable circular orbit around a planet, expressed as \ ( v_o = \sqrt{\frac{GM}{r}} \). This formula is a core topic in Class 11 Physics (NCERT Chapter 8 — Gravitation) and appears frequently in CBSE board exams, JEE Main, and NEET. In this article, we cover the complete derivation, a full formula sheet, three progressive solved examples, common mistakes, and JEE/NEET application strategies.
Key Orbital Velocity Formulas at a Glance
Quick reference for the most important orbital velocity formulas.
- Orbital velocity: \( v_o = \sqrt{\frac{GM}{r}} \)
- Near-surface orbital velocity: \( v_o = \sqrt{gR} \)
- Orbital velocity in terms of height: \( v_o = \sqrt{\frac{GM}{R+h}} \)
- Escape velocity relation: \( v_e = \sqrt{2} \, v_o \) (at same radius)
- Time period of orbit: \( T = \frac{2\pi r}{v_o} \)
- Orbital KE: \( KE = \frac{GMm}{2r} \)
- Total orbital energy: \( E = -\frac{GMm}{2r} \)
What is the Orbital Velocity Formula?
The Orbital Velocity Formula defines the precise speed at which a satellite or celestial body must travel to remain in a stable circular orbit at a given distance from the centre of a planet. At this speed, the gravitational pull of the planet exactly provides the centripetal force required to keep the satellite on its circular path.
According to NCERT Class 11 Physics, Chapter 8 (Gravitation), orbital velocity is derived by equating gravitational force with centripetal force. If the speed is too low, the satellite spirals inward. If the speed is too high, the satellite escapes the orbit entirely.
The formula is applicable to natural satellites (like the Moon), artificial satellites (like the ISS), and even planets orbiting the Sun. Understanding this formula helps students grasp Kepler’s laws, satellite communication, and space mission planning. The Orbital Velocity Formula is one of the most frequently tested concepts in CBSE Class 11 board exams and competitive entrance tests like JEE and NEET.
Orbital Velocity Formula — Expression and Variables
The standard Orbital Velocity Formula for a circular orbit is:
\[ v_o = \sqrt{\frac{GM}{r}} \]
For a satellite orbiting at height h above Earth’s surface, where R is Earth’s radius:
\[ v_o = \sqrt{\frac{GM}{R+h}} \]
For a near-Earth orbit (where \( h \ll R \)), the formula simplifies to:
\[ v_o = \sqrt{gR} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( v_o \) | Orbital velocity | m/s |
| \( G \) | Universal gravitational constant | N·m²/kg² |
| \( M \) | Mass of the central body (planet/star) | kg |
| \( r \) | Orbital radius (from centre of planet) | m |
| \( R \) | Radius of the planet | m |
| \( h \) | Height of orbit above surface | m |
| \( g \) | Acceleration due to gravity at surface | m/s² |
Derivation of the Orbital Velocity Formula
The derivation balances gravitational force with centripetal force. For a satellite of mass \( m \) orbiting at radius \( r \):
Step 1: Write gravitational force: \( F_g = \frac{GMm}{r^2} \)
Step 2: Write centripetal force needed: \( F_c = \frac{mv_o^2}{r} \)
Step 3: Set them equal: \( \frac{GMm}{r^2} = \frac{mv_o^2}{r} \)
Step 4: Cancel \( m \) and simplify: \( v_o^2 = \frac{GM}{r} \)
Step 5: Take the square root: \( v_o = \sqrt{\frac{GM}{r}} \)
Notice that the orbital velocity is independent of the mass of the satellite. Only the mass of the central body and the orbital radius matter.
Complete Gravitation Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Orbital Velocity (general) | \( v_o = \sqrt{\frac{GM}{r}} \) | G = gravitational constant, M = planet mass, r = orbital radius | m/s | Class 11, Ch 8 |
| Orbital Velocity (near surface) | \( v_o = \sqrt{gR} \) | g = surface gravity, R = planet radius | m/s | Class 11, Ch 8 |
| Orbital Velocity (at height h) | \( v_o = \sqrt{\frac{GM}{R+h}} \) | R = planet radius, h = height above surface | m/s | Class 11, Ch 8 |
| Escape Velocity | \( v_e = \sqrt{\frac{2GM}{R}} \) | G = gravitational constant, M = planet mass, R = radius | m/s | Class 11, Ch 8 |
| Relation: Escape and Orbital Velocity | \( v_e = \sqrt{2} \, v_o \) | At the same orbital radius | — | Class 11, Ch 8 |
| Time Period of Orbit | \( T = \frac{2\pi r}{v_o} = 2\pi\sqrt{\frac{r^3}{GM}} \) | r = orbital radius, G = gravitational constant, M = mass | s | Class 11, Ch 8 |
| Orbital Kinetic Energy | \( KE = \frac{GMm}{2r} \) | m = satellite mass, r = orbital radius | J | Class 11, Ch 8 |
| Orbital Potential Energy | \( PE = -\frac{GMm}{r} \) | m = satellite mass, r = orbital radius | J | Class 11, Ch 8 |
| Total Orbital Energy | \( E = -\frac{GMm}{2r} \) | m = satellite mass, r = orbital radius | J | Class 11, Ch 8 |
| Universal Law of Gravitation | \( F = \frac{GMm}{r^2} \) | F = force, M and m = masses, r = separation | N | Class 11, Ch 8 |
| Variation of g with height | \( g_h = \frac{g}{\left(1+\frac{h}{R}\right)^2} \) | g = surface gravity, h = height, R = planet radius | m/s² | Class 11, Ch 8 |
Orbital Velocity Formula — Solved Examples
Example 1 (Class 9-10 Level): Near-Earth Orbital Velocity
Problem: Calculate the orbital velocity of a satellite orbiting very close to Earth’s surface. Take g = 9.8 m/s² and radius of Earth R = 6.4 × 10⁶ m.
Given: g = 9.8 m/s², R = 6.4 × 10⁶ m, h ≈ 0 (near-surface orbit)
Step 1: Use the near-surface orbital velocity formula: \( v_o = \sqrt{gR} \)
Step 2: Substitute the values: \( v_o = \sqrt{9.8 \times 6.4 \times 10^6} \)
Step 3: Multiply inside the root: \( v_o = \sqrt{6.272 \times 10^7} \)
Step 4: Calculate the square root: \( v_o \approx 7920 \) m/s \( \approx 7.92 \) km/s
Answer
The orbital velocity of a near-Earth satellite is approximately 7.92 km/s (≈ 8 km/s).
Example 2 (Class 11-12 Level): Satellite at Given Height
Problem: A satellite orbits Earth at a height of 3.6 × 10⁶ m above the surface. Find its orbital velocity. Take G = 6.67 × 10⁻¹¹ N·m²/kg², mass of Earth M = 6 × 10²⁴ kg, and radius of Earth R = 6.4 × 10⁶ m.
Given: h = 3.6 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg², M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m
Step 1: Find the orbital radius: \( r = R + h = 6.4 \times 10^6 + 3.6 \times 10^6 = 10 \times 10^6 = 10^7 \) m
Step 2: Write the formula: \( v_o = \sqrt{\frac{GM}{r}} \)
Step 3: Substitute values: \( v_o = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{10^7}} \)
Step 4: Simplify the numerator: \( GM = 6.67 \times 6 \times 10^{13} = 40.02 \times 10^{13} \)
Step 5: Divide by r: \( \frac{40.02 \times 10^{13}}{10^7} = 40.02 \times 10^6 \)
Step 6: Take the square root: \( v_o = \sqrt{40.02 \times 10^6} \approx 6326 \) m/s \( \approx 6.33 \) km/s
Answer
The orbital velocity of the satellite at height 3.6 × 10⁶ m is approximately 6.33 km/s.
Example 3 (JEE/NEET Level): Energy and Velocity in Orbit
Problem: A satellite of mass 500 kg orbits Earth at an orbital radius of 8 × 10⁶ m. Find (a) the orbital velocity, (b) the kinetic energy, and (c) the total mechanical energy of the satellite. Take GM = 4 × 10¹⁴ N·m²/kg.
Given: m = 500 kg, r = 8 × 10⁶ m, GM = 4 × 10¹⁴ N·m²/kg
Step 1 — Orbital Velocity: \( v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{4 \times 10^{14}}{8 \times 10^6}} = \sqrt{5 \times 10^7} \approx 7071 \) m/s
Step 2 — Kinetic Energy: \( KE = \frac{GMm}{2r} = \frac{4 \times 10^{14} \times 500}{2 \times 8 \times 10^6} \)
Step 3 — Simplify KE: \( KE = \frac{2 \times 10^{17}}{1.6 \times 10^7} = 1.25 \times 10^{10} \) J
Step 4 — Total Energy: \( E = -\frac{GMm}{2r} = -KE = -1.25 \times 10^{10} \) J
Note: The total energy is always negative for a bound orbit. This confirms the satellite is gravitationally bound to Earth.
Answer
(a) Orbital velocity ≈ 7071 m/s | (b) Kinetic energy = 1.25 × 10¹⁰ J | (c) Total energy = −1.25 × 10¹⁰ J
CBSE Exam Tips 2025-26
- Memorise both forms: Know \( v_o = \sqrt{GM/r} \) for general orbits and \( v_o = \sqrt{gR} \) for near-surface orbits. CBSE often asks you to switch between them.
- Always find r first: In problems where height h is given, always compute \( r = R + h \) before substituting. Forgetting this step is the most common error in board exams.
- State independence from satellite mass: A 2-mark question often asks whether orbital velocity depends on the satellite’s mass. The answer is no — always justify this from the derivation.
- Link to escape velocity: We recommend memorising \( v_e = \sqrt{2} \, v_o \). CBSE 2025-26 papers frequently test this ratio in one-mark and two-mark questions.
- Use standard values: For Earth, use G = 6.67 × 10⁻¹¹ N·m²/kg², M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m, and g = 9.8 m/s² unless the question specifies otherwise.
- Show units in every step: CBSE awards method marks. Always write the unit alongside each calculated value to avoid losing marks on presentation.
Common Mistakes to Avoid
- Using radius instead of orbital radius: Many students substitute R (planet radius) directly when the satellite is at height h. Always use \( r = R + h \) as the orbital radius.
- Confusing orbital velocity with escape velocity: Orbital velocity is \( \sqrt{GM/r} \) while escape velocity is \( \sqrt{2GM/r} \). They differ by a factor of \( \sqrt{2} \). Mixing them up costs marks in both CBSE and JEE.
- Thinking orbital velocity depends on satellite mass: The mass of the satellite cancels out during derivation. Orbital velocity depends only on the central body’s mass and the orbital radius.
- Wrong unit conversion: Orbital radius is often given in kilometres. Always convert to metres before substituting into the formula.
- Forgetting the negative sign for total energy: Total mechanical energy of a satellite in orbit is always negative (\( E = -GMm/2r \)). A positive total energy means the object is unbound and not in orbit.
JEE/NEET Application of the Orbital Velocity Formula
In our experience, JEE aspirants encounter the Orbital Velocity Formula in at least one question per paper, often combined with energy conservation or Kepler’s laws. NEET tests it primarily in the context of satellite communication and Earth’s gravitational field.
Pattern 1: Transfer Orbits and Energy Change
JEE Advanced frequently asks about the energy required to move a satellite from one orbit to another. The change in total energy is:
\[ \Delta E = -\frac{GMm}{2r_2} – \left(-\frac{GMm}{2r_1}\right) = \frac{GMm}{2}\left(\frac{1}{r_1} – \frac{1}{r_2}\right) \]
If \( r_2 > r_1 \), the energy change is positive — you must supply energy to raise the orbit. This is a classic JEE problem type.
Pattern 2: Comparing Orbital Velocities of Planets
NEET and JEE Main often ask students to compare orbital velocities of two satellites at different radii. Since \( v_o \propto \frac{1}{\sqrt{r}} \), a satellite at a greater radius has a smaller orbital velocity. This inverse square-root relationship is a key concept.
Pattern 3: Geostationary Satellite Problems
A geostationary satellite has a time period T = 24 hours. JEE problems ask you to find its orbital radius using \( T = 2\pi\sqrt{r^3/GM} \) and then its orbital velocity. Our experts suggest practising this two-step calculation until it becomes automatic.
For NEET, focus on the conceptual aspect: a geostationary satellite appears stationary because its orbital period matches Earth’s rotation period. It always orbits above the equator.
FAQs on Orbital Velocity Formula
Explore More Physics Formulas
Now that you have mastered the Orbital Velocity Formula, strengthen your Physics preparation with these related topics. Study the Angular Displacement Formula to understand rotational motion, which complements circular orbital mechanics. Review the Spring Constant Formula for oscillatory systems that share mathematical parallels with orbital period equations. Explore the Flow Rate Formula for fluid dynamics applications. For the complete list of Class 11 Physics formulas, visit our Physics Formulas hub. You can also verify the NCERT syllabus for Gravitation directly on the official NCERT website.