Acceleration due to gravity is minimum at

Options

A. The surface of earth
B. The depth 100 km from the earth`s surface
C. The centre of earth
D. The height 100 km from the earth`s surface

Answer

To determine where the acceleration due to gravity is minimum, we can analyze the options provided using the formulas related to gravity at different heights and depths.

### Step-by-Step Solution:

1. **Understanding the Concept**:
- The acceleration due to gravity (g) varies with height above the Earth's surface and depth below the Earth's surface.
- At the surface of the Earth, g is at its maximum value.

2. **Using the Formula for Height**:
- The formula for acceleration due to gravity at a height (h) above the Earth's surface is given by:
\[ g_h = g \left( \frac{R^2}{(R + h)^2} \right) \]
- Here, R is the radius of the Earth (approximately 6400 km), and g is the acceleration due to gravity at the surface.

3. **Using the Formula for Depth**:
- The formula for acceleration due to gravity at a depth (d) below the Earth's surface is given by:
\[ g_d = g \left( 1 - \frac{d}{R} \right) \]
- At the center of the Earth, d equals R, which leads to:
\[ g_{center} = g \left( 1 - 1 \right) = 0 \]

4. **Evaluating Each Option**:
- **Option A**: At the surface of the Earth, g = g (maximum).
- **Option B**: At a depth of 100 km, using the formula:
\[ g_{100km} = g \left( 1 - \frac{100}{6400} \right) = g \left( 1 - \frac{1}{64} \right) = g \left( \frac{63}{64} \right) \]
This value is less than g but still positive.
- **Option C**: At the center of the Earth, g = 0 (minimum).
- **Option D**: At a height of 100 km, using the formula:
\[ g_{100km} = g \left( \frac{6400^2}{(6400 + 100)^2} \right) \]
This value will also be less than g but greater than 0.

5. **Conclusion**:
- The minimum value of acceleration due to gravity occurs at the center of the Earth, where g = 0.

### Final Answer:
The acceleration due to gravity is minimum at the **center of the Earth**.

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