For a satellite in circular orbit at altitude h: v = √(G M / (R + h)). At Earth's surface (h ≈ 0): v_orb ≈ √(g R) ≈ 7.9 km/s. Note v_orb = v_e / √2.
-- NCERT Class 11 Physics, Ch. 7, p. 137Orbital Velocity
8 MCQs9 revision cards9-step worked example
Source: NCERT GravitationOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Orbital Velocity, explained for NEET
Orbital velocity — the speed that keeps a satellite falling around Earth, not into it.
The high-frequency trap in orbital velocity problems: forgetting to include the orbital kinetic energy when computing the energy needed to place a satellite in orbit. Students compute only the gravitational PE change from surface to orbit altitude, losing the KE component — and losing 4 marks.
Core derivation. For a satellite of mass m in circular orbit at radius r = R + h around Earth (mass M), gravity provides the centripetal force:
GMm/r² = mv²/r → v = √(GM/r) = √(GM/(R + h))
This is the orbital velocity (NCERT Class 11 Physics, Chapter 7, page 138). At the surface (h = 0), v₀ = √(GM/R) = √(gR) ≈ 7.9 km/s for Earth.
Key relationships:
- Orbital velocity decreases as orbit radius increases: v ∝ 1/√r.
- Orbital velocity is independent of the satellite's mass — a 500 kg satellite and a 5000 kg satellite at the same altitude orbit at the same speed.
- Relation to escape velocity: v_e = √2 · v_orbital (at the same radius).
Energy in orbit. The total mechanical energy of a satellite in circular orbit is E = −GMm/[2(R + h)]. This equals −KE. The negative sign confirms the orbit is bound. The kinetic energy is KE = GMm/[2(R + h)], and PE = −GMm/(R + h), so PE = 2E and KE = −E.
Watch-out for NEET: When a problem asks "energy required to launch a satellite to orbit at height h," you must compute ΔE = E_orbit − E_surface, which includes both the PE change AND the orbital KE the satellite must acquire. Omitting the KE component is a common error that produces a wrong distractor.
Can you answer these Orbital Velocity MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The orbital velocity of a satellite in a circular orbit close to Earth's surface is approximately:
Show answer and why every option is right or wrong
Answer: C. C is correct. For a near-surface orbit, v₀ = √(gR) = √(9.8 × 6.4 × 10⁶) ≈ 7.9 km/s (NCERT Class 11 Physics, Chapter 7, page 138).
Why A is wrong: A (3.1 km/s) is too low — this is roughly the first cosmic velocity divided by √(2π), a meaningless quantity. The derivation v = √(gR) gives ~7.9 km/s, not ~3 km/s.
Why B is wrong: B (11.2 km/s) is the escape velocity from Earth's surface, not the orbital velocity. Escape velocity = √2 × orbital velocity.
Why D is wrong: D (16.7 km/s) is higher than even the escape velocity (11.2 km/s). No standard orbital or escape speed from Earth's surface matches this value.
MCQ 2Easy RecallPractice
The orbital velocity of a satellite depends on:
Show answer and why every option is right or wrong
Answer: D. D is correct. From v = √(GM/(R + h)), orbital velocity depends on the central body's mass M and the orbital radius (R + h). The satellite's mass m cancels during the derivation (NCERT Class 11 Physics, Chapter 7, page 138).
Why A is wrong: A is wrong because the satellite's mass cancels when equating gravitational force to centripetal force: GMm/r² = mv²/r. The m drops out, leaving v independent of satellite mass.
Why B is wrong: B is wrong because orbital velocity clearly depends on the orbit radius: v = √(GM/r). Larger r gives smaller v.
Why C is wrong: C is wrong for the same reason — satellite mass cancels in the derivation. Orbital velocity has no dependence on m.
MCQ 3Easy RecallPractice
For a satellite in a stable circular orbit, the relationship between orbital velocity (v₀) and escape velocity (vₑ) at the same orbital radius is:
Show answer and why every option is right or wrong
Answer: B. B is correct. At radius r: v₀ = √(GM/r) and vₑ = √(2GM/r) = √2 · √(GM/r) = √2 · v₀.
Why A is wrong: A is wrong. Escape velocity is greater than orbital velocity by a factor of √2, not equal. If vₑ = v₀, the satellite would not have enough energy to escape to infinity.
Why C is wrong: C is wrong. The factor is √2 ≈ 1.414, not 2. Confusing √2 with 2 is an arithmetic slip.
Why D is wrong: D is wrong — this inverts the relationship. Escape velocity is larger than orbital velocity, not smaller.
MCQ 4Direct ApplicationPractice
A satellite orbits Earth in a circular orbit of radius 2R, where R is Earth's radius. If the orbital velocity near the surface is v₀, the orbital velocity at radius 2R is:
Show answer and why every option is right or wrong
Answer: A. A is correct. v = √(GM/r). At the surface, v₀ = √(GM/R). At radius 2R, v = √(GM/2R) = v₀/√2.
Why B is wrong: B is wrong. This assumes v ∝ 1/r (inverse proportionality), but the correct relation is v ∝ 1/√r. Halving would require quadrupling the radius, not doubling.
Why C is wrong: C is wrong. Orbital velocity is not constant — it decreases with increasing orbit radius as v ∝ 1/√r. At double the radius, v must be less than v₀.
Why D is wrong: D is wrong. Orbital velocity decreases with increasing radius, not increases. v ∝ 1/√r means larger orbits have smaller speeds.
MCQ 5Direct ApplicationPractice
Two satellites of masses m and 4m orbit Earth at the same altitude h. The ratio of their orbital velocities is:
Show answer and why every option is right or wrong
Answer: C. C is correct. Orbital velocity v = √(GM/(R + h)) is independent of satellite mass. Both satellites at the same altitude have identical orbital velocities, so the ratio is 1 : 1.
Why A is wrong: A is wrong. This assumes the heavier satellite orbits faster (v ∝ m), but satellite mass cancels in the derivation of orbital velocity. Mass of the satellite does not appear in v = √(GM/r).
Why B is wrong: B is wrong. This assumes v ∝ √m, giving 1 : √4 = 1 : 2. But satellite mass does not affect orbital velocity at all.
Why D is wrong: D is wrong. This assumes the lighter satellite orbits faster, which is also incorrect. Orbital velocity at a given altitude is the same regardless of satellite mass.
MCQ 6Direct ApplicationPractice
If the radius of Earth were doubled while its mass remained unchanged, the orbital velocity of a near-surface satellite would become:
Show answer and why every option is right or wrong
Answer: D. D is correct. Near-surface orbital velocity v₀ = √(GM/R). If R → 2R with M unchanged: v₀' = √(GM/2R) = v₀/√2.
Why A is wrong: A is wrong. This assumes v ∝ 1/R (inverse), but the correct dependence is v ∝ 1/√R. Doubling R reduces v by a factor of √2, not 2.
Why B is wrong: B is wrong. Orbital velocity decreases when orbital radius increases. Doubling is in the wrong direction and the wrong proportionality.
Why C is wrong: C is wrong. Increasing the radius decreases orbital velocity (v ∝ 1/√R), not increases it. √2 times the original would require the radius to be halved.
MCQ 7CalculationPractice
A satellite of mass m is in a circular orbit at height h = R above Earth's surface (R = Earth's radius, M = Earth's mass). The total energy required to move this satellite from the surface of Earth and place it in this orbit is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Energy on the surface: E_surface = −GMm/R (PE only, at rest). Energy in orbit at r = 2R: E_orbit = −GMm/(2 · 2R) = −GMm/(4R). Energy required = E_orbit − E_surface = −GMm/(4R) − (−GMm/R) = −GMm/(4R) + GMm/R = 3GMm/(4R). This accounts for both the PE change and the orbital KE (NCERT Class 11 Physics, Chapter 7, page 140).
Why B is wrong: B (GMm/4R) represents only the magnitude of the satellite's total orbital energy, ignoring the initial surface PE. The correct calculation requires the full difference: E_orbit − E_surface, not just |E_orbit|.
Why C is wrong: C (GMm/2R) is the KE of a satellite orbiting at radius R (not 2R). This confuses the orbit radius with Earth's radius and does not compute the correct energy difference.
Why D is wrong: D (GMm/R) equals the magnitude of the surface PE. This ignores the fact that the satellite retains negative energy in orbit; the required energy is less than |PE_surface|.
MCQ 8CalculationPractice
A satellite is orbiting Earth in a circular orbit of radius r. If its orbital velocity is increased by a factor of √2 (while at radius r), the satellite will:
Show answer and why every option is right or wrong
Answer: B. B is correct. At radius r, v_orbital = √(GM/r) and v_escape = √(2GM/r) = √2 · v_orbital. Increasing orbital velocity by a factor of √2 gives the satellite exactly the escape velocity at that radius, so it escapes to infinity.
Why A is wrong: A is wrong. The satellite is already at orbital velocity for radius r. Any increase in speed means the gravitational force at r is no longer sufficient for circular motion — the orbit cannot remain unchanged.
Why C is wrong: C is wrong. A slight increase in velocity would produce an elliptical orbit with a higher apogee. But increasing by a factor of √2 reaches escape velocity, which means the satellite has enough energy to reach infinity — it does not settle into any bound orbit.
Why D is wrong: D is wrong. Increasing the satellite's speed adds energy, pushing the orbit outward. Falling toward Earth would require reducing the satellite's speed, not increasing it.
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Orbital Velocity: quick recall before you leave
How do you solve a Orbital Velocity question? A worked example
- 1
Given
• m = 200 kg (satellite mass)• h = R = 6.4 × 10⁶ m (orbit altitude equals Earth's radius)• R = 6.4 × 10⁶ m (Earth's radius)• M = 6.0 × 10²⁴ kg (Earth's mass)• G = 6.67 × 10⁻¹¹ N·m²/kg²
- 2
Required
Minimum energy to move the satellite from rest on the surface to a circular orbit at height h = R (orbit radius = 2R).
- 3
Concept
The energy required equals the difference in total mechanical energy between the final orbit and the initial surface position. On the surface, the satellite is at rest, so its energy is purely gravitational PE. In orbit, the total energy includes both KE (from orbital motion) and PE.
- 4
Formulas
• Surface energy: E_surface = −GMm/R (PE only; KE = 0 at rest)• Orbital energy at radius r = 2R: E_orbit = −GMm/(2 · 2R) = −GMm/(4R)• Energy required: ΔE = E_orbit − E_surface
- 5
Substitution
E_surface = −GMm/R = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200) / (6.4 × 10⁶)
E_orbit = −GMm/(4R) = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200) / (4 × 6.4 × 10⁶) - 6
Calculation
First compute GMm = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 200 = 6.67 × 6.0 × 200 × 10⁻¹¹⁺²⁴ = 8.004 × 10¹⁶ J·m
Note: The factor 200 (satellite mass) and the 4 in the denominator (from 2R orbit radius substitution) are exact counting/defined integers and do not limit significant figures. The final answer is governed by the 3-significant-figure precision of G and M.
E_surface = −8.004 × 10¹⁵ / (6.4 × 10⁶) = −1.251 × 10¹⁰ J
E_orbit = −8.004 × 10¹⁵ / (2.56 × 10⁷) = −3.127 × 10⁹ J
ΔE = E_orbit − E_surface = (−3.127 × 10⁹) − (−1.251 × 10¹⁰) = 9.38 × 10⁹ J - 7
Final answer
ΔE ≈ 9.38 × 10⁹ J ≈ 9.38 GJ
Equivalently, in terms of symbols: ΔE = −GMm/(4R) + GMm/R = 3GMm/(4R). - 8
Common trap
Computing only the PE change: ΔPE = −GMm/(2R) − (−GMm/R) = GMm/(2R). This gives ~6.25 × 10⁹ J — significantly less than the correct answer because it ignores the kinetic energy the satellite must acquire to maintain orbital velocity at radius 2R. The orbital KE = GMm/(4R) must be added.
- 9
Similar NEET-style question
A satellite of mass m is launched from Earth's surface to a circular orbit at height h = 2R. Express the minimum energy required in terms of G, M, m, and R. (Answer: 5GMm/(6R) — same method, different orbit radius.)
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What to remember before solving Orbital Velocity questions
Which Orbital Velocity formulas do you need for NEET?
1 formula — click to collapse
Orbital velocity for circular orbit
Speed of circular orbit at altitude h above body of mass M, radius R.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | orbital speed | m/s |
| M | central mass | kg |
| R+h | orbit radius | m |
Valid when
- Circular orbit
- M >> orbiting mass
More in Gravitation: 4 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
Orbital Velocity questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Orbital Velocity?
1 recurring pattern from past papers — click to collapse
Energy required to launch satellite to altitude. Compute change in total mechanical energy.
Multi StepMedium
Common distractors
forgets orbital KE component
Computes only PE change, ignoring orbital KE
Sources
NCERT refs: Class 11 Physics Chapter 7, p.138
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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