Question

Starting from the center of the Earth having radius $ R $, the variation of $ g $ (acceleration due to gravity) is shown by

• A.
• B.
• C.
• D.

Hint:

Inside the Earth, gravity increases linearly with distance from the center, while outside the Earth, gravity decreases according to the inverse square law.

Answer 1

The Correct Option is C

To understand how acceleration due to gravity changes with depth and height relative to Earth's surface, we analyze two key scenarios.

1. Acceleration at Depth (g_depth):
The acceleration due to gravity at depth $d$ below Earth's surface is given by:

$ g_{\text{depth}} = g_{\text{surface}} \left( 1 - \frac{d}{R} \right) $
where:
- $g_{\text{surface}}$ = surface gravity (9.8 m/s²)
- $R$ = Earth's radius
- $d$ = depth below surface

This can alternatively be expressed as:

$ g_{\text{depth}} = g_{\text{surface}} \left( \frac{r}{R} \right) $
where $r = R - d$ is the distance from Earth's center.

2. Acceleration at Height (g_height):
For height $h$ above the surface:

$ g_{\text{height}} = g_{\text{surface}} \left( \frac{R^2}{(R + h)^2} \right) $

3. Key Observations:
a) At Earth's center ($d = R$):
$ g_{\text{depth}} = 0 $ (gravity cancels out)

b) At surface ($d = 0$):
$ g_{\text{depth}} = g_{\text{surface}} $ (maximum value)

c) Above surface ($h > 0$):
Gravity decreases with height as the denominator $(R + h)^2$ grows.

4. Conclusion:
Gravity increases linearly from center to surface, then decreases inversely with square of distance above surface. The governing equations are:

For depth: $ g_{\text{depth}} = g_{\text{surface}} \left( 1 - \frac{d}{R} \right) $
For height: $ g_{\text{height}} = g_{\text{surface}} \left( \frac{R^2}{(R + h)^2} \right) $

Answer 2

The variation of acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth is governed by the following principles:
1. Inside the Earth (i.e., \( r \leq R \)), the acceleration due to gravity \( g \) increases linearly with \( r \), as \( g(r) = g_0 \cdot \frac{r}{R} \), where \( g_0 \) is the acceleration due to gravity at the surface of the Earth. This means as we move from the center of the Earth towards the surface, \( g \) increases linearly.
2. At the surface (\( r = R \)), the acceleration due to gravity reaches its maximum value \( g_0 \).
3. Outside the Earth (i.e., \( r>R \)), the acceleration due to gravity decreases with the square of the distance, following the inverse square law: \( g(r) = \frac{g_0 R^2}{r^2} \).
Thus, when we start from the center and move towards the surface, the gravitational acceleration increases linearly. After reaching the surface, as we move away from the Earth, the gravitational force decreases according to the inverse square law. Therefore, the correct graph representing this variation is a graph that increases and then decreases.