Grav Potential Energy

8 MCQs2 revision cards9-step worked example

Source: NCERT GravitationPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Gravitational potential energy (GPE) is the energy stored in a two-body system due to their gravitational interaction. The formula is U = −GMm/r, where r is the centre-to-centre separation. The negative sign is not optional decoration — it encodes the physics: you must do positive work against gravity to pull masses apart, so the bound state sits below the zero-energy reference at infinity.

NCERT Class 11 Physics Chapter 7 (page 8) defines this with reference U = 0 at r → ∞. This convention is universal in NEET problems. If a question silently uses a different reference (say, U = 0 at the surface), the entire answer shifts — but NEET virtually never does this. Trust the infinity convention unless the stem explicitly states otherwise.

The high-frequency confusion: students mix up gravitational potential energy U = −GMm/r (a property of the two-body system, in joules) with gravitational potential V = −GM/r (a field property of the source mass alone, in J/kg). The formulas differ by one factor of m. When a stem asks "potential energy of the system," use U. When it asks "gravitational potential at a point," use V. Misreading this costs 5 marks (4 lost + 1 negative).

Sign discipline matters. The PE at the surface (r = R) is more negative than at a higher orbit (r = R + h). Moving a mass from the surface to a higher altitude means U becomes less negative — the system gains energy. This is why you must supply energy to lift a satellite. The change in PE is ΔU = −GMm/(R+h) − (−GMm/R) = GMm[1/R − 1/(R+h)], which is positive — energy added to the system.

For problems involving energy conservation near Earth (launch problems, escape problems), you will combine U = −GMm/r with kinetic energy. The total mechanical energy E = KE + U determines whether an orbit is bound (E < 0) or unbound (E ≥ 0).

Practice 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 gravitational potential energy of a two-mass system is defined as U = −GMm/r. What does the negative sign physically indicate?

MCQ 2Easy RecallPractice

At what separation does the gravitational potential energy U = −GMm/r equal zero?

MCQ 3Easy RecallPractice

Gravitational potential energy U = −GMm/r is a property of:

MCQ 4Direct ApplicationPractice

Two identical spheres, each of mass 5.0 kg, are separated by a centre-to-centre distance of 0.50 m. What is the gravitational potential energy of the system? (G = 6.67 × 10⁻¹¹ N·m²/kg²)

MCQ 5Direct ApplicationPractice

A body of mass m is at the Earth's surface. How much energy must be supplied to move it to a height equal to Earth's radius R above the surface? (Express in terms of g, m, and R.)

MCQ 6Direct ApplicationPractice

Gravitational potential energy of a body of mass m at the surface of a planet of mass M and radius R is U₁. If the planet's mass were doubled but its radius remained the same, the new GPE would be:

MCQ 7Concept TrapPractice

A student claims: "The gravitational potential at a point 2R above Earth's surface is −gR/3, and the gravitational potential energy of a mass m at that point is −mgR/3." Which part of the claim is correct?

MCQ 8CalculationPractice

Three identical particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The gravitational potential energy of the system is:

Quick recall before you leave

Worked Example

1
Given
- Mass of satellite: m = 200 kg - Earth's radius: R = 6.4 × 10⁶ m - Altitude of orbit: h = R, so orbital radius r = 2R - Surface gravity: g = 9.8 m/s²
2
Required
Change in gravitational PE: ΔU = U_orbit − U_surface
3
Concept
Gravitational PE for a mass m at distance r from Earth's centre: U = −GMm/r. Since GM = gR², we can write U = −mgR²/r.
4
Formula
U = −mgR²/r At surface (r = R): U_surface = −mgR²/R = −mgR At orbit (r = 2R): U_orbit = −mgR²/(2R) = −mgR/2 ΔU = U_orbit − U_surface = (−mgR/2) − (−mgR) = mgR/2
5
Substitution
ΔU = (200)(9.8)(6.4 × 10⁶) / 2
6
Calculation
Numerator: 200 × 9.8 = 1960; 1960 × 6.4 × 10⁶ = 1.2544 × 10¹⁰ ΔU = 1.2544 × 10¹⁰ / 2 = 6.272 × 10⁹ J ≈ 6.3 × 10⁹ J **Note on exact values:** The factor of 2 in the denominator (from r = 2R) and 200 kg are exact counting/defined numbers and do not limit significant figures. The result is reported to 2 significant figures, governed by g = 9.8 (2 sig figs).
7
Final answer
ΔU ≈ 6.3 × 10⁹ J (positive — energy must be supplied to lift the satellite). This is only the PE change. To actually place the satellite in a circular orbit at this altitude, additional kinetic energy (orbital KE) must be supplied — the total launch energy exceeds ΔU alone.
8
Common trap
Using ΔU = mgh with h = R gives 200 × 9.8 × 6.4 × 10⁶ = 1.25 × 10¹⁰ J — exactly double the correct answer. The formula mgh assumes uniform g, which fails badly when h is comparable to R. Always use U = −GMm/r for large-altitude problems.
9
Similar NEET-style question
A body of mass 10 kg rests on Earth's surface. Find the energy required to move it to a height of 2R above the surface. (Answer: ΔU = mgR × 2/3. At r = 3R, U = −mgR/3. ΔU = −mgR/3 − (−mgR) = 2mgR/3.) ---

Before solving, remember these

Formula

Gravitational potential energy

U = -G M m / r (taking U → 0 as r → ∞). Negative sign reflects that gravity is attractive — work must be done against it to separate masses. For two-body system at separation r.

-- NCERT Class 11 Physics, Ch. 7, p. 8

Formulas

8 formulas — click to collapse

Escape velocity from a body's surface

v_{e} = \frac{2 GM}{R} = 2 g R

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

Gravitational potential energy (point masses)

U = - \frac{GM m}{r}

PE of two-body system; negative because gravity is attractive (work to separate them is positive).

Symbol	Quantity	SI Unit
U	grav PE	J
M, m	two masses	kg
r	separation	m

Valid when

Reference U=0 at r=infinity
Point or spherical masses

g variation with altitude

g_{h} = g (\frac{R}{R + h})^{2} \approx g (1 - 2 h / R)

Gravitational acceleration decreases with altitude above Earth's surface.

Symbol	Quantity	SI Unit
g_h	g at height h	m/s^2
g	surface g	m/s^2
R	Earth radius	m
h	altitude	m

Valid when

Static observer at altitude
Earth treated as uniform sphere

g variation with depth

g_{d} = g (1 - d / R)

Inside Earth (uniform density), g decreases linearly with depth, vanishing at centre.

Symbol	Quantity	SI Unit
g_d	g at depth	m/s^2
g	surface g	m/s^2
R	Earth radius	m
d	depth	m

Valid when

Earth treated as uniform density sphere

Kepler's third law

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

Square of orbital period proportional to cube of semi-major axis. Holds for elliptic orbits about a central mass.

Symbol	Quantity	SI Unit
T	orbital period	s
a	semi-major axis	m
M	central mass	kg

Valid when

Two-body system with central mass M >> orbiting mass
Bound orbit

Orbital velocity for circular orbit

v = \frac{GM}{R + h}

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

Satellite total mechanical energy

E = - \frac{GM m}{2 ( R + h )}

Total energy = KE + PE = -KE (virial). Always negative for bound orbit; E -> 0 at infinity.

Symbol	Quantity	SI Unit
E	total energy	J
M, m	central mass and satellite mass	kg
R+h	orbit radius	m

Valid when

Circular orbit
Bound (E < 0)

Newton's law of gravitation

F = \frac{G m _{1} m _{2}}{r ^{2}}

Attractive force between any two masses. Inverse-square central force.

Symbol	Quantity	SI Unit
F	force	N
G	grav constant = 6.674e-11	N*m^2/kg^2
m1, m2	masses	kg
r	centre-to-centre distance	m

Valid when

Point masses or spherically symmetric distributions
r > sum of body radii (else use shell theorem)

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

4 items — click to collapse

Trap

g at altitude: linear vs inverse-square

Category: Similar Terms

Student treats g(h) as linear in h. Actual: g(R/(R+h))² (inverse-square).

When it triggers

Question asks for g at significant altitude (e.g. R/2 above surface).

How to avoid

Use g_h = g (R/(R+h))². Linear approximation g(1 - 2h/R) only valid for h << R.

Trap

Kepler's third law: linear vs 3/2 power

Category: Similar Terms

Student treats T proportional to a (linear) instead of a^(3/2).

When it triggers

Question gives change in semi-major axis and asks for new period.

How to avoid

T² ∝ a³, so T ∝ a^(3/2). Doubling a multiplies T by 2^(3/2) ≈ 2.83.

Mistake

Treats g(h) as linear in h for significant heights.

Root cause: formula misuse

Correction

Use g_h = g(R/(R+h))² (inverse-square). Linear approximation g(1-2h/R) is only valid for h << R. For h = R/2, the exact formula gives g_h = (2/3)² g = 4g/9, not g(1-1) = 0.

Mistake

Treats T proportional to a (linear) instead of a^(3/2) in Kepler's third law.

Root cause: formula misuse

Correction

T² ∝ a³, so T ∝ a^(3/2). Doubling a multiplies T by 2^(3/2) ≈ 2.83. Quadrupling a → 8× T.

Past Year Questions

8 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

1124 earth days

288 earth days

3225 earth days

4172 earth days

NTA Answer: Option 2(final)

NEET 2025

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :

136 N

216 N

327 N

432 N

NTA Answer: Option 3(final)

NEET 2024Revised key

1 The mass of a planet is th that of the earth and its diameter is half that of the earth. The acceleration due 10 to gravity on that planet is:

119.6 m s–2

29.8 m s–2

34.9 m s–2

43.92 m s–2

NTA Answer: Option 4(revised_final)

NEET 2024Revised key

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is: 5GmM

16R 2GmM

23R GmM

32R GmM

43R

NTA Answer: Option 1(revised_final)

NEET 2023

Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G = gravitational constant) 12Gm 16Gm

1−

2− R R 20Gm 8Gm

3−

4− R R

NTA Answer: Option 2(final)

NEET 2023

A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is 3π the universal constant of gravitation, the quantity represents Gd

1T2

2T3

NTA Answer: Option 1(final)

NEET 2022

A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is

1180 N/kg

20.05 N/kg

350 N/kg

420 N/kg

NTA Answer: Option 3(final)

NEET 2021

The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is

14v

32v

43v

NTA Answer: Option 1(final)

How NEET usually asks this

4 recurring patterns 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

Body weighs W on surface; find weight at height h above surface. Use g_h = g(R/(R+h))^2.

Direct ApplicationEasy

Common distractors

uses linear decrease with h

Treats g as linear in h instead of inverse-square

Given period T and semi-major axis a of one planet, find for another with different a. T^2 ∝ a^3.

Direct ApplicationEasy

Common distractors

uses linear relation

Treats T proportional to a not a^(3/2)

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

Test yourself on this topic with real past-paper questions:

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Grav Potential Keplers Laws

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Grav Potential Energy

Lesson

Practice MCQs

Quick recall before you leave

Gravitational PE (point masses)

Gravitational potential (field quantity)

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Free NEET study resources