Escape Velocity Ratio for Planets of Different Radius and Density
Physics · Gravitation · NEET
Escape velocity does not depend on the escaping body's mass. When comparing two planets, use ve = √(2GM/R). Writing M in terms of density gives the shortcut ve ∝ R√ρ (radius times square-root of density). Memory hook: "Radius pulls full, density pulls its root" — so ve = k·R·√ρ. Pick the form (mass-based ve ∝ √(M/R) or density-based ve ∝ R√ρ) that matches what the question keeps fixed.
Escape velocity scales as R√ρ. Doubling radius multiplies ve by 2 and doubling density multiplies it by √2, so the total factor is 2√2 ≈ 2.83 (the NEET 2016 case), not 4.
Your doubts, answered
Why is escape velocity proportional to R√ρ and not just to R?
Start from ve = √(2GM/R). A planet's mass is M = density × volume = ρ·(4/3)πR³. Substitute: ve = √(2G·ρ·(4/3)πR³ / R) = √((8πGρ/3)·R²) = R·√(8πGρ/3). The constant √(8πG/3) is the same for every planet, so ve ∝ R√ρ. The R² inside the root becomes R outside, and ρ stays inside as √ρ. That is why radius appears to the first power but density only to the half power.
Escape velocity does not contain the mass of the escaping body. Which mass matters?
Only the mass M of the PLANET matters, never the mass m of the rocket or particle. In ½mve² = GMm/R the small mass m cancels from both sides, giving ve = √(2GM/R). So a feather and a spaceship need the same escape speed from the same planet. In ratio problems, never plug the body's mass — only the planet's radius, mass, or density.
When do I use ve ∝ √(M/R) and when ve ∝ R√ρ?
Use whichever quantities the problem holds constant. If the question gives or fixes MASS and RADIUS, use ve ∝ √(M/R). If it gives or fixes RADIUS and DENSITY, use ve ∝ R√ρ. Example: 'equal mass, radius halved' → use the mass form, ve ∝ 1/√R. 'same density, radius 4 times' → use the density form, ve ∝ R, so ve becomes 4 times. Choosing the wrong form forces extra algebra and invites mistakes.
If radius and density both double, does escape velocity just double?
No. Both factors act together. ve ∝ R√ρ. Radius doubling multiplies ve by 2. Density doubling multiplies ve by √2. Total factor = 2 × √2 = 2√2 ≈ 2.83, not 2 and not 4. This is the exact NEET 2016 answer (1 : 2√2). Handle each factor separately, then multiply.
⚠️ The NEET trap ✗ Radius and density both double, so students say escape velocity becomes 2 × 2 = 4 times (treating density the same way as radius). ✓ ve ∝ R√ρ. Radius doubling gives ×2; density doubling gives only ×√2. So ve becomes 2√2 ≈ 2.83 times, giving ratio ve(earth) : ve(planet) = 1 : 2√2. 🧠 Density enters under the square root — double density adds only √2, never a full 2.
Real NEET questions
NEET 2021
The escape velocity from the earth's surface is v. The escape velocity from the surface of another planet of radius four times that of earth and the same mean density is:
A · 3v
B · 4v ✓
C · v
D · 2v
Solution: Use the density form because density is fixed. ve ∝ R√ρ. Here ρ is the same for both planets, so ve ∝ R. Radius becomes 4 times, therefore escape velocity becomes 4v. Answer: B (4v). Trap to avoid: do NOT bring in mass separately — with fixed density the mass grows as R³, and the density form already accounts for it.
NEET 2016 (Phase 1)
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice that of earth is:
A · 1 : 2
B · 1 : 2√2 ✓
C · 1 : 4
D · 1 : √2
Solution: ve ∝ R√ρ. For the planet, R → 2R and ρ → 2ρ. Factor from radius = 2. Factor from density = √2. Combined factor = 2 × √2 = 2√2. So vp = 2√2 · ve, giving ve : vp = 1 : 2√2. Answer: B. Common mistake: multiplying 2 × 2 = 4 by forgetting density is under the root.
ReNEET 2026
Two planets P1 and P2 with equal mass have radii R1 and R2 respectively, where R2 = R1/2. The escape speeds of P1 and P2 are v1 and v2 respectively. Then v2/v1 is:
A · 1/√2
B · 1
C · √2 ✓
D · 2
Solution: Mass is equal for both, so use the mass form: ve = √(2GM/R), i.e. ve ∝ 1/√R. Then v2/v1 = √(R1/R2). With R2 = R1/2, R1/R2 = 2, so v2/v1 = √2. Answer: C. Note: because the question fixes MASS (not density), you must use ve ∝ 1/√R, not ve ∝ R√ρ.
Solved Gravitation NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for escape velocity in terms of density?
ve = R·√(8πGρ/3), which means ve ∝ R√ρ. It comes from substituting M = ρ·(4/3)πR³ into ve = √(2GM/R).
Does escape velocity depend on the mass of the object being launched?
No. The launched body's mass cancels out. Escape velocity depends only on the planet's mass and radius (or radius and density). A pebble and a rocket have the same escape speed from Earth (about 11.2 km/s).
If two planets have the same density, how do their escape velocities compare?
With equal density, ve ∝ R. So escape velocity is directly proportional to radius. A planet with 4 times Earth's radius (same density) has 4 times Earth's escape velocity.
Why does radius appear to the first power but density only to the half power?
Because M ∝ R³, so 2GM/R ∝ R². Taking the square root turns R² into R (first power) while density, which sits inside the root as ρ, becomes √ρ.
Which is the safer relation to memorise for NEET ratio problems?
Memorise both: ve ∝ √(M/R) and ve ∝ R√ρ. Use the mass form when mass and radius are given, and the density form when radius and density are given. Matching the form to the fixed quantities avoids extra steps.
