Connection of Potential Energy Formula with mgh

Potential Energy at Earth's Surface

When an object of mass m is at Earth's surface (distance r = R from Earth's center):

U₁ = \[-\frac{GMm}{R}\]

Here, G is the gravitational constant, M is Earth's mass, and R is Earth's radius.

Potential Energy at Height h

When the same object is lifted to a height h above the surface (distance r = R + h from the center):

U₂ = \[-\frac{GMm}{R+h}\]

Calculate Change in Potential Energy

The increase in potential energy (work done to lift the object) is:

ΔU = U₂ − U₁

ΔU = \[-\frac{GMm}{R+h}-\left(-\frac{GMm}{R}\right)\]

ΔU = GMm\[\left(\frac{1}{R}-\frac{1}{R+h}\right)\]

ΔU = GMm\[\begin{pmatrix}
\frac{R+h-R}{R(R+h)}
\end{pmatrix}\]

ΔU = \[\frac{GMmh}{R(R+h)}\]

Apply Surface Gravity Relationship

At Earth's surface, acceleration due to gravity is related to Earth's mass and radius by:

GM = gR²

Substituting this into our equation:

ΔU = \[\frac{gR^2\cdot mh}{R(R+h)}\]

ΔU = \[\frac{mgh\cdot R}{R+h}\]

Apply the Approximation for Small Heights

Key Condition: When h << R (height is much smaller than Earth's radius)

We can approximate: R + h ≈ R

Therefore:

ΔU = mgh

This is the familiar mgh formula for potential energy!

Approximation is Valid:

The simplified mgh formula is accurate when:

• h is small compared to R: For heights up to a few kilometers above Earth's surface
• Everyday situations: Heights we encounter in daily life (buildings, mountains, airplane altitude)
• School-level problems: Most Class 12 physics problems use this approximation

The approximation breaks down when:

• h is comparable to R: At very high altitudes (thousands of kilometers)
• Satellite calculations: For objects orbiting Earth
• Space exploration: When dealing with interplanetary distances