If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of

Q: Why can't I just use $\Delta U = mgh$ here?

The formula $\Delta U = mgh$ assumes that the gravitational acceleration $g$ remains constant throughout the motion. When you rise to a height equal to the Earth's radius, the gravitational force drops to a quarter of its surface value. Because the force is not constant, you must integrate or use the general potential energy formula.

Q: Why is gravitational potential energy negative?

By convention, we define gravitational potential energy to be zero at an infinite distance where the gravitational pull is zero. Since gravity is an attractive force, you must do work to pull masses apart. Moving closer to the Earth decreases the potential energy, making it negative.

Q: How does $\Delta U = \frac{mgh}{1 + h/R}$ relate to this problem?

The formula $\Delta U = \frac{mgh}{1 + h/R}$ is a highly useful algebraic rearrangement of the general potential energy difference. If you substitute $h = R$ into this formula, you immediately get $\Delta U = \frac{mgR}{1 + R/R} = \frac{1}{2} mgR$.

Q: What happens to the potential energy if the object goes to infinity?

If the object is raised to infinity ($h \to \infty$), the final potential energy $U_f$ becomes $0$. The gain in potential energy would be $\Delta U = 0 - (- \frac{GMm}{R}) = \frac{GMm}{R} = mgR$. This is the binding energy of the object on the surface.

Q: Does this formula apply to other planets as well?

Yes! The derivation is completely general. For any spherical planet of radius $R_p$ and surface gravity $g_p$, raising a mass $m$ to a height equal to its radius requires a potential energy gain of $\frac{1}{2} m g_p R_p$.

Question

If $g$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the Earth, is

Options:

(A) $\frac{1}{2} mgR$

(B) $2 mgR$

(C) $mgR$

(D) $\frac{1}{4} mgR$

Sigma Percentile · Accepted Answer

### Introduction to Gravitational Potential Wells

Imagine standing on the surface of the Earth.

Every step you take, every jump you make, you are interacting with a massive gravitational field.

For everyday heights—like climbing a flight of stairs or even flying in a commercial airplane—we use the familiar formula for the gain in potential energy:

$$\Delta U = mgh$$

But what happens when we venture beyond our immediate surroundings?

What if we lift an object to a height equal to the **radius of the Earth itself** ($h = R$)?

At such astronomical scales, the assumption of a constant gravitational acceleration $g$ completely breaks down.

Let's dive deep into the physics of gravitational potential wells to understand why this happens and how to calculate the exact energy required.

---

### The Failure of the Flat-Earth Approximation

The formula $\Delta U = mgh$ is built on a simplifying assumption: that the Earth is flat and its gravitational field is uniform.

This is a fantastic approximation when $h$ is just a few meters, or even a few kilometers.

However, the Earth's gravitational field actually obeys Newton's Inverse-Square Law:

$$g(r) = \frac{GM}{r^2}$$

As you rise higher, the distance $r$ from the center of the Earth increases, and the strength of gravity decreases.

At a height $h = R$, the distance from the center becomes $2R$.

At this distance, the local acceleration due to gravity drops to:

$$g' = \frac{GM}{(2R)^2} = \frac{g}{4}$$

Because the gravitational force weakens as you lift the object, you are fighting against a progressively smaller force.

Therefore, the actual energy required to lift the object must be **less** than what the constant-field formula $mgR$ predicts.

Let's calculate exactly how much less.

---

### The Master Potential Energy Equation

To find the exact gain in potential energy, we must use the general expression for gravitational potential energy $U(r)$:

$$U(r) = -\frac{GMm}{r}$$

Here, the negative sign is crucial.

It signifies that the gravitational force is attractive, meaning the system is bound.

As the object is moved further away ($r 	o \infty$), the potential energy increases towards its maximum value of zero.

Let's set up our initial and final states:

1. **Initial State (at the surface):** 
   The distance from the center is $r_i = R$.
   $$U_i = -\frac{GMm}{R}$$

2. **Final State (at height $h = R$):** 
   The distance from the center is $r_f = R + h = 2R$.
   $$U_f = -\frac{GMm}{2R}$$

---

### Calculating the Energy Gain

The gain in potential energy $\Delta U$ is the difference between the final and initial potential energies:

$$\Delta U = U_f - U_i$$

Substituting our expressions:

$$\Delta U = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}ight)$$

Simplifying the double negative:

$$\Delta U = \frac{GMm}{R} - \frac{GMm}{2R}$$

$$\Delta U = \frac{GMm}{2R}$$

---

### Connecting to Surface Gravity $g$

We want our final answer in terms of the acceleration due to gravity $g$ at the Earth's surface.

Recall that:

$$g = \frac{GM}{R^2} \implies GM = gR^2$$

Now, substitute $GM = gR^2$ back into our simplified expression for $\Delta U$:

$$\Delta U = \frac{(gR^2)m}{2R}$$

Canceling one factor of $R$ from the numerator and denominator, we arrive at our final elegant result:

$$\Delta U = \frac{1}{2} mgR$$

This is exactly **half** of the value predicted by the naive $mgh$ formula!

This beautiful result perfectly captures how the weakening of gravity over large distances reduces the energy required to escape the Earth's grasp.

Animated Solution for Physics - Gravitation: If $g$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the Earth, is

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Visualized Solution

Visualizing the Setup

Gravitational Potential Energy Formula

Relating $GM$ to Surface Gravity $g$

Initial Potential Energy $U_{i}$

Final Potential Energy $U_{f}$

Calculating the Gain in Potential Energy $Δ U$

Simplifying the Expression

Substituting $GM = g R^{2}$ for the Final Result

Why can't we use $Δ U = m g h$ ?

The Sigma Insight: Gravitational Potential and Potential Energy

Introduction to Gravitational Potential Wells

The Failure of the Flat-Earth Approximation

The Master Potential Energy Equation

Calculating the Energy Gain

Connecting to Surface Gravity $g$

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