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. 135Gravitational Potential Energy
8 MCQs6 revision cards9-step worked example
Source: NCERT GravitationOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Gravitational Potential Energy, explained for NEET
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 134) 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).
Can you answer these Gravitational Potential Energy 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?
Show answer and why every option is right or wrong
Answer: A. A is correct. The negative sign means the bound state has lower energy than the reference (U = 0 at infinity). Positive work must be done against gravity to separate the masses to infinity. (NCERT Class 11 Physics Chapter 7, page 134.)
Why B is wrong: B is wrong. Gravity is always attractive between masses. The negative sign in U reflects binding energy, not a repulsive force.
Why C is wrong: C is wrong. While it is true that U becomes more negative as r decreases, the negative sign itself indicates the bound-state convention (U = 0 at infinity), not merely a direction of change.
Why D is wrong: D is wrong. G is a positive constant (6.674 × 10⁻¹¹ N·m²/kg²). The negative sign in U = −GMm/r is part of the potential energy formula derived from the attractive nature of gravity.
MCQ 2Easy RecallPractice
At what separation does the gravitational potential energy U = −GMm/r equal zero?
Show answer and why every option is right or wrong
Answer: A. A is correct. By the standard convention used in NCERT and NEET, U = 0 when r → ∞. As masses are brought closer from infinity, U becomes negative. (NCERT Class 11 Physics Chapter 7, page 134.)
Why B is wrong: B is wrong. r = 0 would make U = −GMm/0, which diverges — U is not zero at the centre. (The point-mass formula does not apply inside an extended body.)
Why C is wrong: C is wrong. At the surface, U = −GMm/R, which is a large negative value, not zero.
Why D is wrong: D is wrong. At the Moon's orbital radius, U = −GMm/r_Moon, still negative. The only point where U = 0 is at infinite separation.
MCQ 3Easy RecallPractice
Gravitational potential energy U = −GMm/r is a property of:
Show answer and why every option is right or wrong
Answer: D. D is correct. GPE is a system property — it depends on both M and m and their separation r. Neither mass alone "possesses" the PE; it belongs to the configuration. (NCERT Class 11 Physics Chapter 7, page 134.)
Why A is wrong: A is wrong. U depends on both masses (M and m) and their separation. Removing m from the picture removes the PE entirely.
Why B is wrong: B is wrong. Same reasoning — U = −GMm/r requires both masses. The smaller mass alone has no gravitational PE without a companion.
Why C is wrong: C is wrong. The midpoint field strength is a vector property of the field, not a system energy. U is a scalar belonging to the two-body configuration, independent of any specific field point between them.
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)
Show answer and why every option is right or wrong
Answer: C. C is correct. U = −GMm/r = −(6.67 × 10⁻¹¹)(5.0)(5.0)/0.50 = −(6.67 × 10⁻¹¹ × 25)/0.50 = −(6.67 × 10⁻¹¹ × 50) = −3.34 × 10⁻⁹ J. (NCERT Class 11 Physics Chapter 7, page 134.)
Why A is wrong: A is wrong. This value (−6.67 × 10⁻⁹) results from using m₁ × m₂ = 25 and dividing by r² = 0.25 — the force formula's 1/r² — instead of r = 0.50, doubling the answer.
Why B is wrong: B is wrong. This value (−1.33 × 10⁻⁹) combines two slips: only one mass (5.0 kg) in the numerator and r² = 0.25 in the denominator, confusing the PE formula (1/r) with the force formula (1/r²).
Why D is wrong: D is wrong. This value (−6.67 × 10⁻¹⁰) results from using only one mass (5.0 kg) instead of the product m₁ × m₂ = 25 kg² in the numerator.
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.)
Show answer and why every option is right or wrong
Answer: C. C is correct. Using U = −GMm/r and GM = gR²: at surface, U₁ = −GMm/R = −mgR. At height R, r = 2R, so U₂ = −GMm/2R = −mgR/2. Energy supplied = U₂ − U₁ = −mgR/2 − (−mgR) = mgR/2. (NCERT Class 11 Physics Chapter 7, page 134.)
Why A is wrong: A is wrong. mgR would be the answer if the body were lifted to infinity from the surface (escape energy from surface = mgR). Moving to height R is only halfway to infinity in energy terms — the exact answer is mgR/2. (Trap: using mgh with h = R, which assumes uniform g.)
Why B is wrong: B is wrong. mgR/4 would result from incorrectly computing U₂ at r = 4R (height 3R) or from an algebraic error in the subtraction. At height R, r = 2R, giving ΔU = mgR/2.
Why D is wrong: D is wrong. 2mgR comes from mgh with h taken as the distance from Earth's centre, 2R, instead of the height R — and mgh itself assumes a uniform g. It even exceeds the escape energy from the surface, mgR.
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:
Show answer and why every option is right or wrong
Answer: B. B is correct. U = −GMm/R. Doubling M → U_new = −G(2M)m/R = 2(−GMm/R) = 2U₁. Since U₁ is negative, 2U₁ is more negative — the body is more tightly bound. (NCERT Class 11 Physics Chapter 7, page 134.)
Why A is wrong: A is wrong. U₁/2 would apply if the mass were halved, not doubled. U is directly proportional to M.
Why C is wrong: C is wrong. U₁ (unchanged) would only hold if M were unchanged. Doubling M doubles U.
Why D is wrong: D is wrong. 4U₁ would result from squaring the factor of 2, which has no basis in the formula — U is linearly proportional to M, not quadratically.
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?
Show answer and why every option is right or wrong
Answer: D. D is correct. At distance 3R from centre: V = −GM/3R = −gR²/3R = −gR/3. And U = mV = −mgR/3. Both statements are consistent and correct. The key relationship is U = mV — potential energy equals mass × potential. (NCERT Class 11 Physics Chapter 7, page 134.)
Why A is wrong: A is wrong. If V is correct at −gR/3, then U = mV = m(−gR/3) = −mgR/3 is automatically correct too. You cannot have V correct but U = mV incorrect — they are linked by definition.
Why B is wrong: B is wrong. Same logic in reverse — if U = −mgR/3 is correct for mass m, then V = U/m = −gR/3 must also be correct. One cannot be right without the other.
Why C is wrong: C is wrong. Both values follow directly from the standard formulas V = −GM/r and U = −GMm/r with r = 3R and GM = gR². The computation is straightforward and both are 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:
Show answer and why every option is right or wrong
Answer: B. B is correct. The system has three pairs. Each pair contributes U = −Gm²/a. Total U = 3 × (−Gm²/a) = −3Gm²/a. The number of pairs for n particles is n(n−1)/2 = 3(2)/2 = 3. (NCERT Class 11 Physics Chapter 7, page 134.)
Why A is wrong: A is wrong. −Gm²/a accounts for only one pair. An equilateral triangle of 3 particles has 3 pairs (1–2, 1–3, 2–3), so the total is 3 times this value. (Trap: forgetting to count all pairs.)
Why C is wrong: C is wrong. −2Gm²/a counts only 2 pairs. There are 3 distinct pairs in a 3-particle system — the formula for pair count is n(n−1)/2 = 3.
Why D is wrong: D is wrong. −3Gm²/(2a) introduces a spurious factor of 1/2. This error typically comes from confusing the pair-counting formula n(n−1)/2 with dividing U itself by 2. The 1/2 is already used in counting pairs — it does not appear again in the energy sum.
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Gravitational Potential Energy: quick recall before you leave
How do you solve a Gravitational Potential Energy question? A 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.)
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What to remember before solving Gravitational Potential Energy questions
Which Gravitational Potential Energy formulas do you need for NEET?
1 formula — click to collapse
Gravitational potential energy (point masses)
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
More in Gravitation: 4 exam traps and mistakes · 7 formulas · 4 question patterns from its other lessons.
Gravitational Potential Energy questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
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