Escape Velocity NEET 2026: Half-Radius Ratio Explained

What is the correct answer to the escape velocity NEET 2026 question?

The escape velocity NEET 2026 answer is option C: halving the radius at fixed mass increases escape speed by a factor of square root of two. The source is NEET UG 2026, held on 21 June 2026, Code-50. Medium difficulty is the question bank’s classification, not an official NTA rating.

Planet P₂ has the same mass as P₁ but half its radius. Find how many times P₁’s surface escape speed is needed on P₂.

Equal planet masses: $M_{1} = M_{2} = M$
Radius of the second planet:

R_{2} = \frac{R_{1}}{2}

Required ratio:

\frac{v_{2}}{v_{1}}

The options are:

This is a numerical proportionality problem, not a spatial arrangement. No diagram is needed.

How do we derive the escape-speed formula from energy?

At minimum escape speed, the object’s total mechanical energy is zero. This is the NCERT Class 11 Gravitation energy method followed by the official solution. The object approaches infinity with zero speed, where gravitational potential energy is taken as zero.

The symbols distinguish the planet’s mass from the escaping object’s mass:

G & : gravitational constant M & : planet's mass R & : planet's radius m & : escaping object's mass

Ignore air resistance and planetary rotation. At the surface, the kinetic and gravitational potential energies are:

K = \frac{1}{2} m v_{e}^{2}

$U = - \frac{G M m}{R}$ Conservation of mechanical energy gives:

\frac{1}{2} m v_{e}^{2} - \frac{G M m}{R} = 0

\frac{1}{2} m v_{e}^{2} = \frac{G M m}{R}

v_{e}^{2} = \frac{2 G M}{R}

v_{e} = \sqrt{\frac{2 G M}{R}}

The escaping object’s mass cancels, so escape speed does not depend on that mass. The planet’s mass remains in the formula.

How does substituting half the radius give option C?

The smaller radius doubles the quantity inside the square root, not the escape speed itself. Since the planets have equal mass, use the same planet mass in both formulas:

v_{1} = \sqrt{\frac{2 G M}{R_{1}}}

v_{2} = \sqrt{\frac{2 G M}{R_{2}}}

Substitute the smaller radius explicitly:

v_{2} = \sqrt{\frac{2 G M}{R_{1} / 2}}

Dividing by half the original radius multiplies the numerator by two:

\frac{2 G M}{R_{1} / 2} = \frac{2 G M \times 2}{R_{1}} = \frac{4 G M}{R_{1}}

Continue the official substitution sequence:

v_{2} = \sqrt{\frac{4 G M}{R_{1}}} = \sqrt{2 \times \frac{2 G M}{R_{1}}} = \sqrt{2} \times \sqrt{\frac{2 G M}{R_{1}}} = \sqrt{2} v_{1}

Divide by the first planet’s escape speed to obtain option C:

\frac{v_{2}}{v_{1}} = \sqrt{2}

Use the compact equal-mass ratio as a check, not as a replacement for the derivation:

\frac{v_{2}}{v_{1}} = \sqrt{\frac{R_{1}}{R_{2}}} = \sqrt{2}

At fixed mass, a smaller surface radius means a greater escape speed. The requested ratio must therefore exceed one.

The shortcut applies only when planet mass is fixed:

v_{e} \propto \frac{1}{\sqrt{R}} when M is fixed

Why does option D give two instead of square root of two?

Option D reports a speed-squared ratio as a speed ratio. Two is a correct intermediate result, but it is not the requested ratio:

\frac{v_{2}^{2}}{v_{1}^{2}} = \frac{R_{1}}{R_{2}} = 2

Reporting this as the escape-speed ratio drops the final square root. The correction is:

{(\frac{v_{2}}{v_{1}})}^{2} = 2

\frac{v_{2}}{v_{1}} = \sqrt{2}

Choose the positive root because escape speeds are positive. “Escape speed squared doubles” is true; “escape speed doubles” is false. Doubling speed would make its square four times as large.

Before matching an option, check whether your expression represents speed, speed squared, or a ratio of those quantities.

What if the radii are equal but the second planet has four times the mass?

At equal radii, escape speed is proportional to the square root of planet mass. This first original Gravitation practice question changes the fixed quantity. It is not another verified NEET PYQ.

Question: Two planets have equal radii, but the second planet has four times the mass of the first. Find the second planet’s escape speed divided by the first’s.

R_{2} = R_{1}, M_{2} = 4 M_{1}

Worked answer: Start with the general ratio. Equal radii cancel, leaving only the mass ratio inside the square root:

\frac{v_{2}}{v_{1}} = \sqrt{\frac{M_{2} / R_{2}}{M_{1} / R_{1}}}

\frac{v_{2}}{v_{1}} = \sqrt{\frac{M_{2}}{M_{1}}} = \sqrt{4} = 2

Here, speed really does double. Compare what stays fixed rather than memorising the original answer:

Changing mass at fixed radius:

v_{e} \propto \sqrt{M} at fixed R

Changing radius at fixed mass:

v_{e} \propto \frac{1}{\sqrt{R}} at fixed M

What if the radius is halved but the planets have equal density?

Equal density changes the result because the smaller planet also has less mass. The fixed-mass shortcut no longer applies. This is a second original Gravitation practice question, not a sourced PYQ.

Question: Two spherical planets have equal mean density. The second has half the first planet’s radius. Find the second planet’s escape speed divided by the first’s.

ρ_{2} = ρ_{1}, R_{2} = \frac{R_{1}}{2}

Worked answer: Mass is mean density multiplied by volume. At equal mean density, the mass ratio is the cube of the radius ratio:

M = \frac{4}{3} π ρ R^{3}

\frac{M_{2}}{M_{1}} = {(\frac{R_{2}}{R_{1}})}^{3} = {(\frac{1}{2})}^{3} = \frac{1}{8}

Substitute both the mass change and the radius change:

\frac{v_{2}}{v_{1}} = \sqrt{(\frac{M_{2}}{M_{1}}) (\frac{R_{1}}{R_{2}})} = \sqrt{\frac{1}{8} \times 2} = \sqrt{\frac{1}{4}} = \frac{1}{2}

This does not contradict the NEET answer. The original planets had equal mass; the smaller planet here has only one-eighth the mass.

Half the radius at equal mass: escape speed becomes square root of two times the original.
Half the radius at equal density: escape speed becomes one-half the original.

Identify the fixed quantity before choosing a proportionality.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

For a worked example of the same idea, see How to Study Respiration in Plants NEET: Maps, ATP and RQ.

Frequently asked questions

What is the answer to the escape velocity NEET 2026 question?

The answer is option C: √2. For planets of equal mass, escape speed is inversely proportional to the square root of radius, so halving the radius increases escape speed by a factor of √2.

How is the escape velocity formula derived?

Ignoring air resistance and planetary rotation, minimum escape speed corresponds to zero total mechanical energy. Setting surface kinetic energy plus gravitational potential energy to zero gives ½mv² − GMm/R = 0, hence v = √(2GM/R). The escaping object's mass cancels, but the planet's mass remains.

Why doesn't escape velocity double when the radius is halved?

At fixed planet mass, halving the radius doubles escape speed squared, not escape speed. Taking the positive square root gives a speed ratio of √2, which is why option D is incorrect.

What happens to escape velocity if radius is halved at equal density?

For spherical planets of equal mean density, halving the radius reduces the mass to one-eighth. Using v₂/v₁ = √[(M₂/M₁)(R₁/R₂)] gives √[(1/8) × 2] = 1/2. Escape speed is therefore halved, unlike the equal-mass case.