The variation of acceleration due to gravity g with distance d from centre of the Earth is best repr

Q: Why is gravity zero at the center of the Earth?

At the exact center, the mass of the Earth surrounds you equally in all directions. The gravitational pulls from all sides cancel each other out perfectly, resulting in zero net gravity.

Q: Why does gravity increase as we move from the center to the surface?

As you move outwards, the amount of mass 'below' you (closer to the center) increases. Even though you are getting further from the center, the increase in enclosed mass dominates, causing a linear increase in gravity.

Q: Doesn't the mass 'above' us pull us outwards?

According to the Shell Theorem, the gravitational force exerted by a uniform spherical shell on an object inside it is exactly zero. So, the mass further out than your current radius has no net effect on you.

Q: Why does the graph curve downwards outside the Earth?

Once outside, the total mass pulling on you is constant (the whole Earth). The only changing variable is your distance. According to the inverse-square law ($1/d^2$), the force drops off rapidly, creating that specific downward curve.

Q: Is the acceleration due to gravity maximum at the surface?

Yes! At the surface, you have the maximum possible mass pulling you inwards, and you are at the minimum possible distance to experience the pull of the entire Earth.

Question

The variation of acceleration due to gravity $g$ with distance $d$ from centre of the Earth is best represented by ($R =$ Earth's radius)

Sigma Percentile · Accepted Answer

The problem asks us to visualize the journey of gravity from the very center of the Earth out into the vastness of space. It's a classic test of understanding how mass distribution affects gravitational pull.

### Inside the Earth: The Linear Climb

Imagine you are at the exact center of the Earth. The mass of the planet surrounds you equally in all directions. The gravitational pull from the left perfectly cancels the pull from the right, and the pull from above cancels the pull from below. At the center, the net acceleration due to gravity is exactly zero.

As you start moving outwards, towards the surface, something interesting happens. You are moving away from the center, but you are also enclosing more and more mass 'beneath' you. According to the **Shell Theorem**, the outer layers of the Earth (the mass at a radius greater than your current distance) exert no net gravitational force on you. You only feel the pull of the sphere of mass *inside* your current radius.

Mathematically, the acceleration due to gravity $g$ at a distance $d$ (where $d < R$) is given by:

$$g = \frac{GM}{R^3}d$$

Here, $G$ is the universal gravitational constant, $M$ is the total mass of the Earth, and $R$ is its radius. Notice that the term $\frac{GM}{R^3}$ is entirely constant. This means that $g$ is **directly proportional** to the distance $d$:

$$g \propto d$$

Graphically, a direct proportionality represents a **straight line passing through the origin**. As you move from the center to the surface, gravity increases linearly, reaching its maximum value exactly at the surface ($d = R$).

### Outside the Earth: The Inverse Square Drop

Now, imagine you have breached the surface and are traveling into space ($d > R$). The situation changes completely. You are no longer enclosing more mass; the total mass pulling on you is now the constant mass of the entire Earth, $M$.

From the outside, the Earth behaves gravitationally as if all its mass were concentrated at a single point at its center. The formula for gravity now follows Newton's familiar law of universal gravitation:

$$g = \frac{GM}{d^2}$$

In this region, $g$ is **inversely proportional** to the square of the distance $d$:

$$g \propto \frac{1}{d^2}$$

This is the famous **inverse-square law**. It means that as you double your distance from the center, the gravity doesn't just halve; it drops to one-fourth of its value. Graphically, this relationship is represented by a curve that slopes downwards, approaching zero asymptotically as you move infinitely far away.

### The Complete Picture

To find the correct graph, we just need to stitch these two realities together:

1.  From $d = 0$ to $d = R$, the graph must be a **straight line** starting from the origin and sloping upwards.
2.  From $d = R$ onwards, the graph must be a **decreasing curve** that gets flatter as it goes further out.

Looking at the given options, only **Graph (c)** perfectly captures this dual nature of Earth's gravity. It shows the linear ascent through the planet's interior, followed by the inverse-square descent into the cosmos.

Animated Solution for Physics - Gravitation: The variation of acceleration due to gravity $g$ with distance $d$ from centre of the Earth is best represented by ( $R =$ Earth's radius)

Select Answer:

Visualized Solution

$g$ vs $d$ Graph

Inside the Earth ( $d < R$ )

Linear Variation

Outside the Earth ( $d > R$ )

Inverse Square Variation

Final Graph

The Sigma Insight: Acceleration due to Gravity and its Variation

Inside the Earth

Outside the Earth

The Complete Picture

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