JEE Mains Physics · Gravitation

Escape Velocity and Energy Conservation

Escape velocity is √(2GM/R) = √(2gR), independent of the body's mass and direction; any launch or fall over a large distance is solved with KE − GMm/r conserved.

All Gravitation notes JEE Mains Physics strategy

Why this matters

Twenty-three PYQs, twenty-two of them multiple choice, and two from 2026. Twelve scale the escape velocity from one planet to another; eleven use conservation of energy for a launch, a fall or an escape from a height. Four are statement or assertion questions. Whenever g would change over the distance travelled, energy conservation, not the constant-g equations, gives the answer.

Concept 1 of 2: Escape velocity and how it scales between planets

To escape, a body must climb from

-GMm/R

to zero energy, so its kinetic energy must be at least

GMm/R

. The mass m cancels, and energy does not care about direction. What is left depends only on the planet: its mass and radius, or its density and radius.

Definition

$v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR} = R\sqrt{\dfrac{8\pi G\rho}{3}}$ ; 11.2 km/s for the earth.
It does not depend on the mass of the body or on the angle of projection.
Mass and radius given: $v_e \propto \sqrt{M/R}$ . Two planets with the same $M/R$ have the same $v_e$ .
Density and radius given: $v_e \propto R\sqrt{\rho}$ .
g and radius given: $v_e \propto \sqrt{gR}$ .
Near the surface $v_e = \sqrt{2}\,v_o$ , where $v_o = \sqrt{gR}$ is the speed of a grazing orbit.
The moon keeps no atmosphere because its $v_e$ (about 2.4 km/s) is small enough for gas molecules to reach.

Escape velocity

v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = R\sqrt{\frac{8\pi G\rho}{3}}

Worked example

A planet has 50 times the earth's mass and twice its radius. The escape velocity from the earth is 11.2 km/s. Find it for the planet.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q5Moderate

Example 1 · Gravitation · Escape Velocity and Energy Conservation

The escape velocity from a spherical planet A is 10

km/s

. The escape velocity from another planet B whose density and radius are

10\%

of those of planet A, is

\_\_\_\_

m/s

Equal escape velocity needs equal M/R

Two planets of different mass can have the same escape velocity only if M/R is the same. Equal MR, or equal M/R², does not do it.

With density given, v_e grows with R, not √R

v_e = R√(8πGρ/3). Doubling the radius at the same density doubles v_e. Writing √(M/R) and forgetting that M grows as R³ gives √2 instead.

Direction does not matter

A body thrown sideways, at 45° or straight up needs the same speed to escape. Escape is an energy condition, and kinetic energy has no direction.

Drill 11 more on escape velocity and how it scales between planets

Concept 2 of 2: Energy conservation for launches and falls

When a body travels a distance comparable to R, g changes along the way, so

v^{2} = u^{2} + 2gh

is wrong. Use energy instead: kinetic energy plus

-GMm/r

stays constant. Write it at the start and at the end, with r always measured from the centre.

Definition

$\tfrac{1}{2}mv_1^{2} - \dfrac{GMm}{r_1} = \tfrac{1}{2}mv_2^{2} - \dfrac{GMm}{r_2}$ . Use $GM = gR^{2}$ .
Energy needed to escape from the surface: $\dfrac{GMm}{R} = mgR$ .
Escape from a point at distance r from the centre: $v = \sqrt{2GM/r}$ .
Launched with $\lambda v_e$ : the body rises to $r_{\max} = \dfrac{R}{1 - \lambda^{2}}$ from the centre.
Falling from rest at distance r: $v^{2} = 2GM\left(\dfrac{1}{R} - \dfrac{1}{r}\right)$ . From infinity this is $v_e$ .
From a grazing orbit, the extra speed to escape is $(\sqrt{2} - 1)\sqrt{gR}$ .
Between two bodies, a launch only has to reach the neutral point, where the two pulls balance; after that the other body pulls it in.

Mechanical energy is conserved

\tfrac{1}{2}mv_1^{2} - \frac{GMm}{r_1} = \tfrac{1}{2}mv_2^{2} - \frac{GMm}{r_2} \qquad r_{\max} = \frac{R}{1 - \lambda^{2}}

Worked example

A body is released from rest at a height 2R above the earth's surface. Find its speed when it reaches the surface. Take

g = 10\ \text{m/s}^{2}

and

R = 6.4 \times 10^{6}

m; ignore air.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q98Moderate

Example 2 · Gravitation · Escape Velocity and Energy Conservation

A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be: (Take radius of earth

= 6400\text{ }km

and

g = 10{\text{ }ms}^{- 2}

)

A fall from height R gives √(gR), not √(2gR)

v² = 2gh assumes g stays the same over the whole fall. Over a distance R it does not; energy conservation gives v² = gR.

Escape energy is mgR, not ½mgR

The body must climb from −GMm/R to zero, which is GMm/R = mgR. Half of it, ½mgR, is the kinetic energy of a grazing orbit, a different quantity.

r is from the centre

In GMm/r, a point 3R above the surface has r = 4R. Using the height in place of r is the usual slip.

Drill 10 more on energy conservation for launches and falls

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Escape velocity and how it scales between planets
Escape velocity
$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = R\sqrt{\frac{8\pi G\rho}{3}}$
Energy conservation for launches and falls
Mechanical energy is conserved
$\tfrac{1}{2}mv_1^{2} - \frac{GMm}{r_1} = \tfrac{1}{2}mv_2^{2} - \frac{GMm}{r_2} \qquad r_{\max} = \frac{R}{1 - \lambda^{2}}$

Watch out for (6)

Equal escape velocity needs equal M/R
→ Escape velocity and how it scales between planets
With density given, v_e grows with R, not √R
→ Escape velocity and how it scales between planets
Direction does not matter
→ Escape velocity and how it scales between planets
A fall from height R gives √(gR), not √(2gR)
→ Energy conservation for launches and falls
Escape energy is mgR, not ½mgR
→ Energy conservation for launches and falls
r is from the centre
→ Energy conservation for launches and falls

Test yourself on Gravitation

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Gravitation chapter test Drill all 23 questions