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You have used mgh since Class 9, and it works for balls, lifts and water tanks. Then a JEE question asks how much energy it takes to lift a satellite to a height equal to Earth's radius, and mgh gives twice the right answer. This post shows where mgh comes from, when it breaks, and how to handle the problems it cannot: large heights and systems of several masses.
Why is mgh only an approximation?
The force of gravity falls off as 1/r², so g is smaller at greater heights. OpenStax University Physics notes that ΔU = mg(y₂ − y₁) "works very well if g does not change significantly" between the two heights, and then derives the general form. NCERT's Gravitation chapter says the same in its points to ponder: the familiar mgh is "actually an approximation" to the difference in gravitational potential energy.
The general result, with zero potential energy at infinite separation, is
U = −GMm / r
where r is the distance between the centres. It is negative because you would have to do work on the mass to pull it away to infinity, where U is zero.
How does mgh come out of −GMm/r?
Move a mass m from Earth's surface (r = R) to a height h (r = R + h):
ΔU = GMm [1/R − 1/(R + h)] = GMm h / [R(R + h)].
Use GM = gR², where g is the surface value:
ΔU = mgh / (1 + h/R).
When h is tiny compared with R, the bracket is almost 1 and you get mgh back. So mgh overestimates the true value by a factor of (1 + h/R). That gives a quick rule: the error in mgh is about h/R. With R ≈ 6400 km, mgh is within 1% up to about 64 km.
For numbers, this post uses R = 6.4 × 10⁶ m and g = 9.8 m/s². NASA's Earth fact sheet gives a volumetric mean radius of 6371 km and a mean surface gravity of 9.820 m/s², so these rounded values are close.
Worked example 1: lifting a mass to a height equal to Earth's radius
How much energy is needed to raise a 100 kg payload from the surface to a height h = R?
With h = R: ΔU = mgR / (1 + 1) = mgR/2.
ΔU = 100 × 9.8 × 6.4 × 10⁶ / 2 ≈ 3.14 × 10⁹ J.
Using mgh instead: 100 × 9.8 × 6.4 × 10⁶ ≈ 6.27 × 10⁹ J, twice the correct value.
Worked example 2: a modest-looking height
Find the energy per kilogram to lift a mass to 400 km above the surface.
h/R = 400 / 6400 = 0.0625.
mgh per kg: 9.8 × 4 × 10⁵ = 3.92 × 10⁶ J/kg.
Exact: 3.92 × 10⁶ / 1.0625 ≈ 3.69 × 10⁶ J/kg.
Here mgh is about 6% too high. That is enough to pick the wrong option in a multiple-choice question.
Worked example 3: how fast must it be launched?
What vertical launch speed would carry a body up to a height R before it stops (ignoring air)?
Kinetic energy at launch = rise in potential energy: ½mv² = mgR/2.
v = √(gR) = √(9.8 × 6.4 × 10⁶) ≈ 7.9 km/s.
If you had used mgh, you would get v = √(2gR) ≈ 11.2 km/s, which is in fact the speed needed to escape Earth completely. Our post on gravitation and satellites covers escape speed and orbital energy in detail.
How do you find the potential energy of several masses?
Potential energy belongs to pairs. For a system of point masses, NCERT's rule is to add −Gm₁m₂/r for every pair, once each.
Example 4. Three masses, each m, sit at the corners of an equilateral triangle of side a. There are three pairs, each at distance a, so U = −3Gm²/a. The work needed to separate them completely is +3Gm²/a. With m = 10 kg and a = 1 m, that is 3 × 6.67 × 10⁻¹¹ × 100 ≈ 2.0 × 10⁻⁸ J, a reminder of how weak gravity is between everyday objects.
NCERT works the square version: four equal masses at the corners of a square of side l have four side pairs and two diagonal pairs, giving U ≈ −5.41Gm²/l.
What mistakes do students make?
Using mgh at large heights, without checking h/R.
Measuring r from the surface instead of from Earth's centre.
Dropping the minus sign in −GMm/r and getting energy changes with the wrong sign.
Counting pairs twice in a system of masses, or missing the diagonal pairs.
Treating g at height h as the surface value when using GM = gR². The g in that identity is always the surface value.
A practice routine for gravitational energy
Derive ΔU = mgh / (1 + h/R) yourself once, from −GMm/r.
Solve three height problems with h/R = 0.01, 0.1 and 1, and compare each with mgh.
Solve two launch-speed problems by energy conservation.
Solve two system problems: a triangle and a square of masses.
For each answer, ask whether the sign makes sense: lifting should need positive work.
If the jump from mgh to −GMm/r is the part that does not make sense yet, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revisit gravitational potential energy on its own.
Today, work out the h/R error for a mountain, a jet's cruising height and a 400 km orbit, and see where mgh stops being safe.
