Formula
Escape velocity
The minimum speed required to escape Earth's gravity: v_e = √(2 G M / R) = √(2 g R) ≈ 11.2 km/s. Independent of the mass of the projectile; depends only on the gravitating body.
-- NCERT Class 11 Physics, Ch. 7, p. 136Escape Velocity
8 MCQs3 revision cards9-step worked example
Source: NCERT GravitationPYQ coverage: NEET 2021, 2026Official key: NTA-verifiedLast updated: 25 Sep 2026
Escape Velocity, explained for NEET
Escape velocity is the minimum launch speed needed for an object to leave a planet's gravitational field permanently — reaching infinity with zero residual speed. The derivation is energy conservation: set total mechanical energy at launch equal to zero (the boundary between bound and unbound trajectories).
At the surface, KE + PE = 0 gives ½mv² + (−GMm/R) = 0, yielding v_e = √(2GM/R). Since g = GM/R², this simplifies to v_e = √(2gR). For Earth, v_e ≈ 11.2 km/s (NCERT Class 11 Physics Chapter 7, page 136).
Key features you must internalise:
- Independent of the launched object's mass. The m cancels. A marble and a rocket have the same escape velocity from a given planet.
- Independent of launch direction. The derivation uses energy (scalar), not force components. Horizontal, vertical, or at 45° — the threshold speed is identical.
- Depends only on M and R of the planet (or equivalently, g and R). This is the basis of every NEET scaling question.
The high-frequency NEET pattern asks you to compare escape velocities when a planet's radius, mass, or density changes relative to Earth. The relationship v_e = √(2GM/R) can be rewritten using M = (4/3)πR³ρ as v_e = R√(8πGρ/3). This density form is what examiners exploit: when density is given instead of mass, students who substitute M directly get the wrong scaling.
Watch-out: forgetting to convert density-radius data into mass before applying v_e ∝ √(M/R) leads to a common distractor. Always decide first: am I given M directly, or must I express M in terms of ρ and R?
Can you answer these Escape 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 escape velocity from the surface of a planet depends on:
Show answer and why every option is right or wrong
Answer: D. From v_e = √(2GM/R), escape velocity depends only on the planet's mass M and radius R. The escaping object's mass cancels out, and the derivation uses energy (a scalar), so direction is irrelevant (NCERT Class 11 Physics Chapter 7, page 136).
Why A is wrong: A is wrong because v_e = √(2GM/R) shows dependence on the planet's M and R, not the escaping object's mass — the object's mass m cancels in the energy equation.
Why B is wrong: B is wrong because the derivation uses conservation of energy (scalar equation), not force components. Launch direction does not appear in v_e = √(2GM/R).
Why C is wrong: C is wrong because the escaping object's mass m cancels from both sides of ½mv² = GMm/R. Only the planet's M and R remain.
MCQ 2Easy RecallPractice
The escape velocity from the surface of the Earth is approximately:
Show answer and why every option is right or wrong
Answer: A. The standard NCERT value is v_e ≈ 11.2 km/s for Earth (Class 11 Physics Chapter 7, page 136).
Why B is wrong: B is wrong — 7.9 km/s is the orbital velocity for a near-surface satellite (v_o = √(gR)), not escape velocity. A common confusion since v_e = √2 × v_o.
Why C is wrong: C is wrong — 3.1 km/s is too low; it does not correspond to any standard gravitational speed for Earth.
Why D is wrong: D is wrong — 22.4 km/s is exactly 2 × 11.2, suggesting the student doubled v_e instead of recognising that v_e = √2 × v_o, not 2 × v_o.
MCQ 3Easy RecallPractice
The escape velocity from a planet's surface is related to the orbital velocity of a satellite in a near-surface orbit by:
Show answer and why every option is right or wrong
Answer: D. v_o = √(GM/R) and v_e = √(2GM/R), so v_e = √2 × v_o. This is a standard result from NCERT Class 11 Physics Chapter 7.
Why A is wrong: A is wrong because v_e = √(2GM/R) while v_o = √(GM/R). The factor of 2 inside the square root gives v_e = √2 × v_o, not equality.
Why B is wrong: B is wrong — this inverts the relationship. v_e is larger than v_o, not smaller.
Why C is wrong: C is wrong — this gives v_e = 2 × v_o, which would mean v_e² = 4GM/R. The actual relation has v_e² = 2GM/R, so the factor is √2, not 2.
MCQ 4Direct ApplicationPractice
A planet has mass 4M and radius 2R, where M and R are Earth's mass and radius respectively. The escape velocity from this planet's surface, in terms of Earth's escape velocity v_e, is:
Show answer and why every option is right or wrong
Answer: C. v_e(planet) = √(2G × 4M / 2R) = √(2 × 2GM/R) = √2 × √(2GM/R) = √2 × v_e. The mass-to-radius ratio doubles, and √2 appears from the square root (NCERT Class 11 Physics Chapter 7, page 136).
Why A is wrong: A is wrong — this would require M/R to be unchanged, but 4M/(2R) = 2(M/R), not M/R. The escape velocity increases by √2.
Why B is wrong: B is wrong — this result comes from taking √(4) = 2 without accounting for the denominator change. The correct ratio under the root is 4M/(2R) = 2(M/R), giving √2, not 2.
Why D is wrong: D is wrong — this likely comes from multiplying the mass factor (4) by itself or misapplying the formula without the square root. v_e scales as √(M/R).
MCQ 5Direct ApplicationPYQ Pattern
A planet has 4 times the radius and the same average density as Earth. The ratio of escape velocity from this planet to that from Earth is:
Show answer and why every option is right or wrong
Answer: B. Express M in terms of density: M = (4/3)πR³ρ. Then v_e = √(2G(4/3)πR³ρ/R) = R√(8πGρ/3), so v_e ∝ R when density is constant. Planet radius is 4R, so v_e(planet)/v_e(Earth) = 4R/R = 4. The ratio is 4 : 1 (NCERT Class 11 Physics Chapter 7, page 136; pattern from PYQ 2021).
Why A is wrong: A is wrong — this would hold only if both mass and radius were identical. With 4× radius at same density, mass increases as R³ = 64×, and the net effect on √(M/R) is not unity.
Why C is wrong: C is wrong — getting 2 typically results from taking √(4) = 2, which would apply only to v_e ∝ √R. But when density is held constant, v_e ∝ R (first power), not √R. The density form of the formula must be used.
Why D is wrong: D is wrong — 8 would arise from cubing the radius factor (4³ = 64) and mishandling the square root. The correct scaling with constant density is v_e ∝ R, giving a ratio of 4, not 8.
MCQ 6Direct ApplicationPractice
If the mass of Earth were doubled and its radius halved, the new escape velocity would be:
Show answer and why every option is right or wrong
Answer: A. v_e' = √(2G × 2M / (R/2)) = √(2G × 4M/R) = 2 × √(2GM/R) = 2 × v_e. The M/R ratio quadruples (2M ÷ R/2 = 4M/R), and √4 = 2 (NCERT Class 11 Physics Chapter 7, page 136).
Why B is wrong: B is wrong — doubling mass alone would give √2 × v_e, and halving radius alone would also give √2 × v_e. Combined, the effect is 2 × v_e, not unchanged.
Why C is wrong: C is wrong — getting 4 means the student forgot the square root. Under the radical, 2M/(R/2) = 4(M/R), but √4 = 2, not 4.
Why D is wrong: D is wrong — √2 would result from accounting for only one change (either doubling mass OR halving radius) but not both simultaneously.
MCQ 7Concept TrapPractice
An object is launched from a planet's surface at exactly the escape velocity. Which statement about its journey is correct?
Show answer and why every option is right or wrong
Answer: C. At exactly v_e, total mechanical energy E = KE + PE = 0. The object reaches infinity (r → ∞) where PE → 0 and thus KE → 0, meaning it arrives at infinity with zero speed. This is the boundary between bound (E < 0) and unbound (E > 0) trajectories (NCERT Class 11 Physics Chapter 7, page 136).
Why A is wrong: A is wrong — if the object retained v_e at infinity, its total energy would be ½mv_e² > 0, meaning it was launched above escape velocity. At exactly v_e, all KE converts to overcoming gravity.
Why B is wrong: B is wrong — returning to the surface requires E < 0 (bound trajectory). At v_e, E = 0, which is the unbound threshold. The object never returns.
Why D is wrong: D is wrong — a stable circular orbit requires a specific orbital velocity v_o = v_e/√2 and a centripetal condition. Launch at v_e provides too much energy for a bound orbit.
MCQ 8CalculationPractice
A planet has radius R and surface gravitational acceleration g. A body is projected vertically upward from the surface with speed v = v_e / 2, where v_e is the escape velocity. The maximum height reached by the body above the surface is:
Show answer and why every option is right or wrong
Answer: B. Use energy conservation. At the surface: ½m(v_e/2)² − GMm/R = −GMm/(R+h). Since v_e² = 2GM/R, we have ½m(2GM/4R) − GMm/R = −GMm/(R+h). This gives GMm/(4R) − GMm/R = −GMm/(R+h), so −3GMm/(4R) = −GMm/(R+h). Cancel and invert: R+h = 4R/3, giving h = R/3 (NCERT Class 11 Physics Chapter 7).
Why A is wrong: A is wrong — R/4 comes from treating g as constant: ½mv² = mgh with v² = v_e²/4 = gR/2 gives h = R/4. Over a height comparable to R, g falls off, so the body rises higher than this.
Why C is wrong: C is wrong — h = R would require the initial KE to be half the magnitude of the surface PE, but (v_e/2)² gives only 1/4 of v_e², which is 1/4 of 2GM/R = GM/(2R). The energy budget doesn't support h = R.
Why D is wrong: D is wrong — h = 2R would require a launch speed closer to v_e. At v_e/2, only 1/4 of the escape energy is provided, yielding a much smaller maximum height.
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Escape Velocity: quick recall before you leave
How do you solve a Escape Velocity question? A worked example
Pattern: Escape velocity scaling when planet parameters change (NEET pattern: escape velocity scaling, PYQ 2021)
- 1
Given
• Planet radius: R' = 4R_E• Planet density: ρ' = 2ρ_E• Earth escape velocity: v_e (known)
- 2
Required
Escape velocity from the planet's surface, expressed as a multiple of v_e.
- 3
Concept
Escape velocity depends on mass and radius: v_e = √(2GM/R). When density is given, substitute M = (4/3)πR³ρ to get the density form: v_e = R√(8πGρ/3). This reveals v_e ∝ R√ρ.
- 4
Formula
v_e = R√(8πGρ/3), so v_e ∝ R√ρ
- 5
Substitution
v_e(planet) / v_e(Earth) = (R' × √ρ') / (R_E × √ρ_E) = (4R_E × √(2ρ_E)) / (R_E × √ρ_E)
- 6
Calculation
= 4 × √2 = 4√2
Note: The factors 4 (radius ratio) and 2 (density ratio) are exact problem-defined integers. They do not limit significant figures. The constants 8, π, G, and 3 in the formula are mathematical/physical constants and are also exact for this ratio calculation. - 7
Final answer
v_e(planet) = 4√2 × v_e ≈ 5.66 × v_e
- 8
Common trap
Forgetting to convert density to mass. If a student directly uses v_e ∝ √(M/R) without computing M = (4/3)π(4R)³(2ρ) = 128 × (4/3)πR³ρ = 128M_E, they may apply the wrong scaling. The density form v_e ∝ R√ρ avoids this error.
- 9
Similar NEET-style question
A planet has 3 times the radius and 3 times the average density of Earth. What is the escape velocity from this planet, given that Earth's escape velocity is 11.2 km/s?
Approach: v_e ∝ R√ρ → ratio = 3 × √3 = 3√3 ≈ 5.20. Answer: 5.20 × 11.2 ≈ 58.2 km/s.
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What to remember before solving Escape Velocity questions
Which Escape Velocity formulas do you need for NEET?
1 formula — click to collapse
Escape velocity from a body's surface
Minimum speed for an object to escape gravity to infinity from radius R. Earth: ~11.2 km/s.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_e | escape velocity | m/s |
| M | planet mass | kg |
| R | planet radius | m |
| g | surface gravity | m/s^2 |
Valid when
- Launched from surface
- No air drag
- Body treated as point/sphere
More in Gravitation: 4 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
Escape Velocity questions from past NEET papers
2 questions from NEET 2021, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
11/√2
21
3√2
42
NTA Answer: Option 3(final)
NEET 2021
14v
2v
32v
43v
NTA Answer: Option 1(final)
How does NEET ask about Escape Velocity?
1 recurring pattern from past papers — click to collapse
Compare escape velocities for planets with different M, R. v_e ∝ sqrt(M/R). Common shape: 'planet has 4× radius and 2× density of Earth — find v_e'.
Direct ApplicationEasy
Common distractors
forgets density not mass
Confuses planet mass with density given radius scaling
Sources
NCERT refs: Class 11 Physics Chapter 7, p.136
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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