The Circular Velocity Formula gives the minimum speed an object must maintain to stay in a stable circular orbit around a massive body, expressed as \ ( v_c = \sqrt{\frac{GM}{r}} \). This formula is a core concept in NCERT Class 11 Physics, Chapter 8 — Gravitation. It is equally critical for JEE Main, JEE Advanced, and NEET aspirants who face numericals on orbital mechanics every year. In this article, we cover the full derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET application patterns.
Key Circular Velocity Formulas at a Glance
Quick reference for the most important circular velocity and orbital mechanics formulas.
- Circular orbital velocity: \( v_c = \sqrt{\frac{GM}{r}} \)
- Near-Earth orbital velocity: \( v_c = \sqrt{gR} \)
- Escape velocity: \( v_e = \sqrt{\frac{2GM}{r}} = \sqrt{2} \, v_c \)
- Time period of orbit: \( T = \frac{2\pi r}{v_c} \)
- Angular velocity relation: \( v_c = r\omega \)
- Centripetal acceleration: \( a_c = \frac{v_c^2}{r} = \frac{GM}{r^2} \)
- Orbital radius from surface: \( r = R + h \)
What is Circular Velocity?
The Circular Velocity Formula describes the precise speed at which a satellite or any orbiting body must travel to maintain a perfectly circular path around a central mass. At this speed, the gravitational pull of the central body acts as the centripetal force. The object neither spirals inward nor escapes outward — it remains in a stable orbit.
This concept appears in NCERT Class 11 Physics, Chapter 8 (Gravitation). The NCERT textbook introduces orbital velocity in the context of Earth satellites. Students learn how the Moon, artificial satellites, and space stations maintain their orbits.
Circular velocity depends on two factors: the mass of the central body (M) and the orbital radius (r). Crucially, it does not depend on the mass of the orbiting object. A 1 kg satellite and a 1000 kg satellite at the same orbital radius have identical circular velocities. This is a key conceptual point tested in CBSE board exams and competitive entrance tests.
For a satellite orbiting close to Earth's surface (h ≈ 0), the formula simplifies to \( v_c = \sqrt{gR} \approx 7.9 \) km/s. This value is frequently asked in CBSE numericals.
Circular Velocity Formula — Expression and Variables
The general expression for circular velocity is derived by equating gravitational force to centripetal force:
\[ v_c = \sqrt{\frac{GM}{r}} \]
For a satellite orbiting at height h above a planet of radius R, the orbital radius becomes r = R + h:
\[ v_c = \sqrt{\frac{GM}{R + h}} \]
For a near-surface orbit (h ≈ 0), using g = GM/R²:
\[ v_c = \sqrt{gR} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( v_c \) | Circular (orbital) velocity | m/s |
| \( G \) | Universal gravitational constant | N·m²·kg⁻² |
| \( M \) | Mass of the central body (planet/star) | kg |
| \( r \) | Orbital radius (from centre of central body) | m |
| \( R \) | Radius of the planet | m |
| \( h \) | Height of orbit above planet surface | m |
| \( g \) | Acceleration due to gravity at surface | m/s² |
Derivation of the Circular Velocity Formula
The derivation follows directly from Newton's law of gravitation and the requirement for circular motion.
Step 1: Write the gravitational force on a satellite of mass m at orbital radius r:
\[ F_g = \frac{GMm}{r^2} \]
Step 2: Write the centripetal force needed to keep the satellite in circular orbit:
\[ F_c = \frac{mv_c^2}{r} \]
Step 3: For a stable circular orbit, set gravitational force equal to centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv_c^2}{r} \]
Step 4: Cancel mass m from both sides and simplify:
\[ v_c^2 = \frac{GM}{r} \]
Step 5: Take the square root to get the final expression:
\[ v_c = \sqrt{\frac{GM}{r}} \]
Notice that the mass of the satellite (m) cancels out completely. This confirms that circular velocity is independent of the orbiting object's mass.
Complete Gravitation Formula Sheet
The table below covers all key gravitation formulas from NCERT Class 11, Chapter 8. Use this as your quick revision sheet before board exams and competitive tests.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Circular (Orbital) Velocity | \( v_c = \sqrt{\frac{GM}{r}} \) | G = gravitational constant, M = planet mass, r = orbital radius | m/s | Class 11, Ch 8 |
| Near-Surface Orbital Velocity | \( v_c = \sqrt{gR} \) | g = surface gravity, R = planet radius | m/s | Class 11, Ch 8 |
| Escape Velocity | \( v_e = \sqrt{\frac{2GM}{r}} \) | G, M, r as above | m/s | Class 11, Ch 8 |
| Relation: Escape and Circular Velocity | \( v_e = \sqrt{2} \, v_c \) | v_e = escape velocity, v_c = circular velocity | m/s | Class 11, Ch 8 |
| Time Period of Orbit | \( T = \frac{2\pi r}{v_c} = 2\pi\sqrt{\frac{r^3}{GM}} \) | T = period, r = orbital radius | s | Class 11, Ch 8 |
| Universal Law of Gravitation | \( F = \frac{GMm}{r^2} \) | F = force, M and m = masses, r = distance | N | Class 11, Ch 8 |
| Acceleration due to Gravity | \( g = \frac{GM}{R^2} \) | g = surface gravity, R = planet radius | m/s² | Class 11, Ch 8 |
| Variation of g with Height | \( g_h = g\left(1 – \frac{2h}{R}\right) \) | g_h = gravity at height h, h << R | m/s² | Class 11, Ch 8 |
| Gravitational Potential Energy | \( U = -\frac{GMm}{r} \) | U = potential energy, m = satellite mass | J | Class 11, Ch 8 |
| Orbital Kinetic Energy | \( KE = \frac{1}{2}mv_c^2 = \frac{GMm}{2r} \) | KE = kinetic energy, m = satellite mass | J | Class 11, Ch 8 |
| Total Orbital Energy | \( E = -\frac{GMm}{2r} \) | E = total mechanical energy (negative for bound orbit) | J | Class 11, Ch 8 |
| Angular Velocity Relation | \( v_c = r\omega \) | ω = angular velocity | m/s | Class 11, Ch 7 |
Circular Velocity Formula — Solved Examples
Work through these three examples carefully. They progress from basic CBSE-level substitution to advanced JEE-style reasoning.
Example 1 (Class 9-10 / Basic CBSE Level)
Problem: Calculate the orbital velocity of a satellite revolving 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 circular velocity formula: \( v_c = \sqrt{gR} \)
Step 2: Substitute the values: \( v_c = \sqrt{9.8 \times 6.4 \times 10^6} \)
Step 3: Calculate the product inside the square root: \( 9.8 \times 6.4 \times 10^6 = 62.72 \times 10^6 \)
Step 4: Take the square root: \( v_c = \sqrt{6.272 \times 10^7} \approx 7920 \) m/s
Answer
The orbital velocity of a near-surface satellite is approximately 7.92 km/s ≈ 7.9 km/s. This is the standard value quoted in NCERT textbooks.
Example 2 (Class 11-12 / Multi-Step CBSE Level)
Problem: A satellite orbits Earth at a height of 600 km above the surface. Find its orbital velocity and time period. Take G = 6.674 × 10⁻¹¹ N·m²·kg⁻², mass of Earth M = 6 × 10²⁴ kg, radius of Earth R = 6.4 × 10⁶ m.
Given: h = 600 km = 6 × 10⁵ m, G = 6.674 × 10⁻¹¹ N·m²·kg⁻², M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m
Step 1: Find the orbital radius r = R + h: \( r = 6.4 \times 10^6 + 6 \times 10^5 = 7.0 \times 10^6 \) m
Step 2: Apply the circular velocity formula: \( v_c = \sqrt{\frac{GM}{r}} \)
Step 3: Substitute values: \( v_c = \sqrt{\frac{6.674 \times 10^{-11} \times 6 \times 10^{24}}{7.0 \times 10^6}} \)
Step 4: Simplify the numerator: \( 6.674 \times 6 \times 10^{13} = 40.044 \times 10^{13} \)
Step 5: Divide: \( \frac{40.044 \times 10^{13}}{7.0 \times 10^6} = 5.72 \times 10^7 \)
Step 6: Take the square root: \( v_c = \sqrt{5.72 \times 10^7} \approx 7563 \) m/s ≈ 7.56 km/s
Step 7: Find the time period: \( T = \frac{2\pi r}{v_c} = \frac{2\pi \times 7.0 \times 10^6}{7563} \)
Step 8: Calculate T: \( T = \frac{4.398 \times 10^7}{7563} \approx 5815 \) s ≈ 97 minutes
Answer
Orbital velocity ≈ 7.56 km/s. Time period ≈ 97 minutes. This matches the typical low-Earth orbit parameters of the International Space Station.
Example 3 (JEE/NEET Application Level)
Problem: A satellite is orbiting Earth at radius r₁. It is then boosted to a higher orbit at radius r₂ = 4r₁. Find the ratio of the new orbital velocity to the original orbital velocity. Also find the ratio of the new time period to the original time period.
Given: Initial orbital radius = r₁, Final orbital radius = r₂ = 4r₁
Step 1: Write circular velocity at r₁: \( v_1 = \sqrt{\frac{GM}{r_1}} \)
Step 2: Write circular velocity at r₂ = 4r₁: \( v_2 = \sqrt{\frac{GM}{4r_1}} = \frac{1}{2}\sqrt{\frac{GM}{r_1}} = \frac{v_1}{2} \)
Step 3: Find the velocity ratio: \( \frac{v_2}{v_1} = \frac{1}{2} \)
Step 4: Use the time period formula \( T \propto r^{3/2} \) (Kepler's third law):
Step 5: Find the time period ratio: \( \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} = (4)^{3/2} = 8 \)
Step 6: Alternatively verify using T = 2πr/v: \( \frac{T_2}{T_1} = \frac{r_2/v_2}{r_1/v_1} = \frac{4r_1 \cdot 2}{r_1 \cdot 1} = 8 \) ✓
Answer
The new orbital velocity is half the original (v₂/v₁ = 1/2). The new time period is 8 times the original (T₂/T₁ = 8). This type of ratio problem is very common in JEE Main and NEET.
CBSE Exam Tips 2025-26
- Memorise the two key values: Near-surface orbital velocity ≈ 7.9 km/s and escape velocity ≈ 11.2 km/s. These appear directly in 1-mark and 2-mark questions.
- Always identify r correctly: The orbital radius r is measured from the centre of the planet, not from the surface. Add the planet's radius R to the height h before substituting into the formula.
- Show the derivation step: CBSE 3-mark and 5-mark questions often ask for the derivation. We recommend writing all five steps clearly, starting from equating gravitational and centripetal forces.
- Remember the independence from satellite mass: A common 1-mark question asks whether orbital velocity depends on the satellite's mass. The answer is no — m cancels in the derivation. State this explicitly.
- Use the relation v_e = √2 · v_c: If you know the circular velocity, you can instantly find the escape velocity. This saves time in multi-part questions in the 2025-26 board paper.
- Practice unit conversion: Answers are often expected in km/s. Convert m/s to km/s by dividing by 1000. Forgetting this step costs marks in CBSE board exams.
Common Mistakes to Avoid
Our experts have reviewed thousands of student answer scripts. These are the most frequent errors in circular velocity problems:
- Mistake 1 — Using r as height above surface: Many students substitute h directly into the formula instead of r = R + h. Always add the planet's radius to the given height. This single error accounts for the majority of wrong answers in CBSE board exams.
- Mistake 2 — Confusing circular velocity with escape velocity: Circular velocity is \( \sqrt{GM/r} \) and escape velocity is \( \sqrt{2GM/r} \). They differ by a factor of √2. Never mix them up. Escape velocity is always greater than circular velocity at the same radius.
- Mistake 3 — Including satellite mass in the formula: Some students write \( v_c = \sqrt{GMm/r} \) by mistake. The satellite mass m cancels during the derivation. The correct formula has no m in it.
- Mistake 4 — Wrong value of G: The universal gravitational constant is G = 6.674 × 10⁻¹¹ N·m²·kg⁻². Students sometimes write 6.67 × 10⁻¹⁰ or use wrong units. Always double-check the power of 10.
- Mistake 5 — Forgetting that v_c decreases as r increases: Since \( v_c \propto 1/\sqrt{r} \), a satellite at a higher orbit moves slower, not faster. This is counterintuitive but physically correct. Higher orbit means lower speed and longer time period.
JEE/NEET Application of Circular Velocity Formula
In our experience, JEE aspirants encounter the circular velocity formula in at least 2-3 questions per year, either directly or as part of a larger orbital mechanics problem. NEET typically tests conceptual understanding rather than heavy calculation.
Pattern 1 — Ratio and Comparison Problems (JEE Main)
JEE Main frequently asks students to compare orbital velocities or time periods at different radii. The key relationships to remember are:
- \( v_c \propto \frac{1}{\sqrt{r}} \) — doubling the orbital radius reduces speed by factor √2
- \( T \propto r^{3/2} \) — Kepler's third law connects time period and orbital radius
- \( KE = \frac{GMm}{2r} \), \( PE = -\frac{GMm}{r} \), \( E_{total} = -\frac{GMm}{2r} \) — energy relationships follow from v_c
Pattern 2 — Energy Analysis in Orbital Transfer (JEE Advanced)
JEE Advanced problems often involve transferring a satellite from one circular orbit to another (Hohmann transfer). You must calculate the change in kinetic energy, potential energy, and total energy. All these calculations start from the circular velocity formula. The total energy is always negative for a bound orbit, and it becomes less negative (increases) as the orbit radius increases.
Pattern 3 — Conceptual Questions (NEET)
NEET tests conceptual understanding with questions like: “What happens to orbital speed when the satellite moves to a higher orbit?” or “Which quantity remains unchanged when a satellite is placed in a higher orbit?” The answer patterns are:
- Speed decreases at higher orbit (v_c ∝ 1/√r)
- Time period increases at higher orbit (T ∝ r^{3/2})
- Total energy increases (becomes less negative) at higher orbit
- Angular momentum increases at higher orbit (L = mv_c r ∝ √r)
In our experience, students who clearly understand the derivation of the circular velocity formula can answer all four of these NEET patterns without memorising each result separately.
FAQs on Circular Velocity Formula
Explore More Physics Formulas
Now that you have mastered the Circular Velocity Formula, strengthen your understanding of related topics. Visit our comprehensive Physics Formulas hub for a complete collection of NCERT-aligned formula articles.
For closely related concepts, we recommend reading the Angular Velocity Formula article, which covers the relationship between angular speed and linear speed in circular motion. You should also study the Normal Force Formula to understand how contact forces interact with circular motion problems. For fluid and advanced mechanics, our Fluid Mechanics Formula guide provides further context on force and velocity relationships. Finally, the Instantaneous Speed Formula article will help you connect instantaneous linear speed to orbital mechanics. For the official NCERT syllabus and textbook content, refer to the NCERT official website.