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Exam level4 past questions

Gravitational potential energy and work

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, and work is required to move it against gravity.

Why this shows up in the exam

NEET often asks you to compute energy changes, work done, and their applications to satellites and objects near Earth.

How NEET tests this

Numerical · 4 Qs

Learn the idea

Gravitational potential energy of a mass in Earth’s field is negative and given by U = -GM_E m / r; the change in U equals the work done by an external agent when the mass is moved radially.

🧠 Memory hook: Gravity’s pit: the deeper (smaller r) you go, the more negative the energy – think of falling into a deeper hole.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • U = -GM_E m / r (zero at infinity)
  • g = GM_E / R_E² so GM_E = g R_E²
  • For small heights ΔU ≈ m g Δh
  • Work done by external agent = +ΔU, work done by gravity = -ΔU
  • Always use distance from Earth’s centre r = R_E + height
  • Potential energy becomes less negative as r increases

How to approach it

  1. 1Read the question and note the height above the surface
  2. 2Convert height to distance from centre: r = R_E + h
  3. 3Use U = -g R_E² m / r (or -GM_E m / r) to find the required U or ΔU
  4. 4Apply sign convention: external work = +ΔU
  5. 5Choose the answer that matches the algebraic value

Worked example — watch it click

Potential energy of a satellite having mass 'm' and rotating at a height of 6.4 x 10⁶ m from the surface of Earth is:

  • ✅-0.5 mgR
  • B)-mgR
  • C)-2 mgRₑ
  • D)4 mgR

The concept behind this problem

The example forces you to convert the given height into the true radial distance and then substitute into U = -gR_E² m / r, revealing the -½ mgR_E result.

Step by step

  1. 1The distance of the satellite from Earth’s centre is r = R_(E)+h = 6.4×10⁶ m+6.4×10⁶ m=2R_(E).
  2. 2Gravitational potential energy of a mass m at distance r is U = -G M_(E) mr.
  3. 3Using g = G M_(E)R_(E)² gives G M_(E)=gR_(E)².
  4. 4Substituting, U = -gR_(E)² m2R_(E) = -(1)/(2) m g R_(E).
  5. 5Thus the satellite’s potential energy equals -0.5 mgR_(E).
  6. 6The option -mgR would correspond to the surface (r = R), -2mgR_(E) would imply a radius half the Earth’s, and a positive value (4 mgR) cannot be gravitational potential energy.
  7. 7So the correct answer is -0.5 mgR.

Watch out

Students often forget to add Earth’s radius to the height and take r = h, leading to an incorrect potential energy.

Common slip-ups that cost marks

  • •Using h instead of R_E + h gives a wrong r
  • •Mixing up the sign: forgetting that U is negative and external work is positive when U increases
  • •Applying the near‑Earth approximation mgΔh when the height is comparable to R_E

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 4NEET 2001

Potential energy of a satellite having mass 'm' and rotating at a height of 6.4 x 10⁶ m from the surface of Earth is:

Push further

More challenging

2 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 2

A body of mass 'm' is lifted from the Earth's surface to a height 'h' where h is much smaller than the Earth's radius (h << R). Which of the following expressions best approximates the increase in its gravitational potential energy?

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