What is the fractional decrease in the value of free-fall acceleration g for a particle when it is lifted from the surface to an elevation `h`? `(h lt lt R)`

Explanation

To find the fractional decrease in the value of free-fall acceleration \( g \) when a particle is lifted from the surface to an elevation \( h \) (where \( h \ll R \), with \( R \) being the radius of the Earth), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula for Gravitational Acceleration**:
The acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth is given by:
\[ g = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Differentiating the Equation**:
To find how \( g \) changes with respect to \( r \), we differentiate \( g \) with respect to \( r \):
\[ \frac{dg}{dr} = -\frac{2GM}{r^3} \]

3. **Substituting the Change in Radius**:
When the particle is lifted to a height \( h \), the new radius becomes \( r = R + h \). However, since \( h \ll R \), we can approximate \( r \) as \( R \) for small changes. Thus, we can use \( dr = h \):
\[ \frac{dg}{dh} = -\frac{2GM}{R^3} \]

4. **Relating \( dg \) to \( g \)**:
We know that \( g = \frac{GM}{R^2} \). We can express \( \frac{dg}{g} \) in terms of \( h \):
\[ \frac{dg}{g} = \frac{dg}{dh} \cdot \frac{dh}{g} \]
Substituting \( \frac{dg}{dh} \):
\[ \frac{dg}{g} = -\frac{2GM}{R^3} \cdot \frac{h}{\frac{GM}{R^2}} = -\frac{2h}{R} \]

5. **Finding the Fractional Decrease**:
The fractional decrease in \( g \) is given by:
\[ \frac{dg}{g} = -\frac{2h}{R} \]
Hence, the fractional decrease in the value of \( g \) when lifted to height \( h \) is:
\[ \text{Fractional decrease} = -\frac{2h}{R} \]

### Final Answer:
The fractional decrease in the value of free-fall acceleration \( g \) for a particle when it is lifted from the surface to an elevation \( h \) is:
\[ -\frac{2h}{R} \]