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

Q: Why can't we just use the formula $mgh$ for potential energy?

The formula $mgh$ is an approximation that only works when $h$ is very small compared to the radius of the Earth ($h \ll R$). In this problem, $h = R$, which is huge! Over such large distances, the value of $g$ decreases significantly, so we must use the exact formula $U = -\frac{GMm}{r}$.

Q: Why is the gravitational potential energy negative?

By convention, potential energy is set to zero at infinity. Since gravity is an attractive force, you have to do positive work to move an object from a finite distance to infinity. This means the energy at any finite distance must be less than zero, hence negative.

Q: Why do we measure the distance from the center of the Earth and not the surface?

According to Newton's Shell Theorem, a spherically symmetric body like the Earth exerts a gravitational force on external objects as if all its mass were concentrated at its exact geometric center. Therefore, all radial distances must be measured from the center.

Q: What happens to the kinetic energy of the object?

The question only asks for the *gain in potential energy*, which is the work done against gravity to slowly raise the object. We assume the object starts at rest and ends at rest, so kinetic energy doesn't factor into this specific calculation.

Q: How did we get $GM_e = gR^2$?

The acceleration due to gravity $g$ is the gravitational force per unit mass at the surface. Using Newton's law of gravitation, $F = \frac{GM_e m}{R^2}$. Since $F = mg$, equating the two gives $mg = \frac{GM_e m}{R^2}$, which simplifies to $g = \frac{GM_e}{R^2}$. Multiplying both sides by $R^2$ gives $GM_e = gR^2$.

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) $2 mgR$

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

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

(D) $mgR$

Sigma Percentile · Accepted Answer

### The Setup: Understanding Gravitational Potential Energy

Imagine you are standing on the surface of the Earth, holding a ball of mass $m$. The Earth, with its massive bulk $M_e$ and radius $R$, is pulling this ball downwards.

This invisible pull means the ball possesses **Gravitational Potential Energy**. But here is the catch—in physics, we define the potential energy to be exactly zero at an infinite distance away.

Because gravity is an attractive force, you have to do work against it to move the ball away. This means the potential energy everywhere closer than infinity must be negative!

At the surface of the Earth, the distance from the center is $R$. Therefore, the initial potential energy $U_i$ is given by the master equation:

$$U_i = -\frac{GM_e m}{R}$$

### The Journey: Moving Upwards

Now, let's take this ball on a journey. We raise it to a height $h$ above the surface.

The problem states that this height is exactly equal to the radius of the Earth, so $h = R$.

To find the new potential energy, we must calculate the total distance from the center of the Earth. This new distance $r$ is the radius $R$ plus the height $h$.

$$r = R + R = 2R$$

At this new height, the gravitational grip is weaker, and the potential energy $U_f$ becomes less negative (which means it has increased):

$$U_f = -\frac{GM_e m}{2R}$$

### The Core Calculation: Finding the Gain

The question asks for the **gain** in potential energy. Gain simply means the final value minus the initial value.

Let's set up the raw equation for the change in potential energy $\Delta U$:

$$\Delta U = U_f - U_i$$

Substituting our values into this equation, we get:

$$\Delta U = -\frac{GM_e m}{2R} - \left(-\frac{GM_e m}{R}\right)$$

Watch out for the minus signs! The double negative becomes a positive, flipping the terms around:

$$\Delta U = \frac{GM_e m}{R} - \frac{GM_e m}{2R}$$

If you have one full unit and you subtract a half unit, you are left with exactly one half. So, the expression simplifies beautifully to:

$$\Delta U = \frac{GM_e m}{2R}$$

### The Final Touch: Bringing in 'g'

We have our answer, but it's not in the format of the options. The options are written in terms of $g$, the acceleration due to gravity at the Earth's surface.

Recall the fundamental relation for $g$:

$$g = \frac{GM_e}{R^2}$$

We can rearrange this to isolate the gravitational constant and mass of the Earth:

$$GM_e = gR^2$$

Now, let's substitute this powerful substitution back into our $\Delta U$ equation:

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

One $R$ in the numerator cancels perfectly with the $R$ in the denominator.

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

And there we have it! The gain in potential energy is exactly half of $mgR$. This is a classic JEE concept that beautifully bridges universal gravitation with the local gravity we experience every day.

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

$U_{i}$ at Surface

$U_{f}$ at Height $h = R$

Gain in Potential Energy $Δ U$

Simplifying $Δ U$

Relation with $g$

Final Substitution

The Way Forward

The Sigma Insight: Gravitational Potential and Potential Energy

The Setup

The Journey

The Core Calculation

The Final Touch

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