Concept understanding — Gravitational Potential Energy
Gravitational Potential Energy
The Intuition: Energy Stored by Height
Imagine holding a heavy book above the floor. Your arm feels tired — that's because you're working against gravity. If you let go, the book falls and gains speed. Where did that motion come from? It came from the position of the book. By lifting it, you stored energy in the Earth–book system. That stored energy is gravitational potential energy.
The higher you lift, the more energy you store. The heavier the object, the more energy you store. This is the core idea: Gravitational potential energy is the energy an object has because of its position in a gravitational field.
The Precise Definition
Gravitational potential energy (U) is the work done against gravity to bring an object from a reference point (usually the ground) to its current position.
For objects near the Earth's surface (where gravity is roughly constant), the formula is beautifully simple:
U=mgh
Where:
U = gravitational potential energy (joules, J)
m = mass of the object (kg)
g = acceleration due to gravity (≈ 9.8 m/s² on Earth)
h = height above the reference point (m)
Why "Potential"?
The word "potential" means "stored and ready to be used." The book at height h has the potential to do work — it can smash a table, compress a spring, or generate sound when it hits the ground. That energy was put in when you lifted it.
The Reference Point is Arbitrary
Here's a crucial point: Only changes in gravitational potential energy matter. You can choose any height as h=0. In most problems, we take the ground as zero, but you could take the floor, the tabletop, or even the ceiling.
If you lift a 2 kg book from the floor (h=0) to a shelf (h=2 m), the change in potential energy is:
ΔU=mgΔh=2×9.8×2=39.2 J
If you instead took the shelf as h=0, the book on the floor would have negative potential energy (−39.2 J). The difference between the two positions is still 39.2 J — that's what matters.
Watch out
Never say "the object has mgh energy" without specifying the reference level. The value is meaningless without a zero point.
The Bigger Picture: Variable Gravity
The formula U=mgh works only when g is constant — that is, near Earth's surface. For large distances (like a rocket leaving Earth), gravity weakens with distance. The general formula for gravitational potential energy between two masses M and m separated by distance r is:
U=−rGMm
The negative sign means that potential energy is zero at infinite separation and becomes more negative as objects come closer. This is the true definition, and U=mgh is a special case of it (derived by approximating near the surface).
Key Takeaways for Exams
Gravitational potential energy is always relative — you must state or imply a reference level.
It depends on height, not path — lifting straight up or along a ramp stores the same energy (if friction is ignored).
It converts to kinetic energy when the object falls: mgh=21mv2 (ignoring air resistance).
The formula U=mgh is for near-Earth problems only. For orbital mechanics, use U=−GMm/r.
Tip
In numerical problems, always write ΔU=mgΔh rather than U=mgh — this reminds you that only changes matter and forces you to define your zero level.
A Simple Example
A 5 kg stone is on a cliff 20 m high. Take the cliff base as h=0.
Potential energy of stone: U=5×9.8×20=980 J
If the stone falls, just before hitting the ground, all this becomes kinetic energy: 21×5×v2=980⟹v=392≈19.8 m/s
That's the energy of position, converted to energy of motion.
For quick revision, remember that Gravitational Potential Energy is drawn directly from the Gravitation coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers, which is exactly why "Gravitational Potential Energy important questions" shows up so often in Physics question banks. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.
Over one full year, the Earth returns to essentially the same orbital position, and since gravity is conservative, the net work over any closed path is zero.
✓Final answer
(a) Zero.
Step 1. Work done by a conservative force equals W=−ΔU=U(ri)−U(rf), which depends only on the initial and final positions, not on the path taken.
Step 2. Over one full year, the Earth completes one full orbit and returns to (essentially) the same position relative to the Sun, so ri=rf.
Step 3. Since U depends only on r, U(ri)=U(rf), so W=U(ri)−U(rf)=0.
Step 4. This does not mean the Sun's force did nothing throughout the year -- it did plenty of positive and negative work along the way (positive as Earth approached the Sun, negative as it receded) -- but these all cancel out exactly over one complete closed orbit.
✓Final answer
(a) Zero -- the net work over one complete orbit is zero, since Earth returns to essentially the same position and gravity is conservative.
Recognise the Sun's force is conservative and Earth returns to the same position after one full orbit.
Assuming the work must be nonzero simply because the Sun is 'always pulling' -- the sign of the work reverses over the two halves of the orbit and cancels.