The value of acceleration due to gravity on the surface of earth is `x`. At an altitude of `h` from the surface of the earth, its value is `y`. If `R` is the radius of earth, then the value of `h` is

Options

A. `(sqrt(x/y)-1)R`
B. `(sqrt(y/x)-1)R`
C. `sqrt(y/x)R`
D. `sqrt(x/y)R`

Explanation

To solve the question, we need to find the height \( h \) at which the acceleration due to gravity changes from \( x \) (on the surface of the Earth) to \( y \) (at height \( h \)). The formula for gravitational acceleration at a distance \( r \) from the center of the Earth is given by:

\[ g = \frac{GM}{r^2} \]

Where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( r \) is the distance from the center of the Earth.

1. **Gravitational acceleration on the surface of the Earth:**
\[ x = \frac{GM}{R^2} \quad \text{(where \( R \) is the radius of the Earth)} \]

2. **Gravitational acceleration at height \( h \):**
\[ y = \frac{GM}{(R + h)^2} \]

3. **Setting up the equations:**
We have two equations:
- Equation 1: \( x = \frac{GM}{R^2} \)
- Equation 2: \( y = \frac{GM}{(R + h)^2} \)

4. **Dividing Equation 1 by Equation 2:**
\[ \frac{x}{y} = \frac{(R + h)^2}{R^2} \]

5. **Cross-multiplying:**
\[ x \cdot R^2 = y \cdot (R + h)^2 \]

6. **Taking the square root of both sides:**
\[ \sqrt{\frac{x}{y}} = \frac{R + h}{R} \]

7. **Rearranging the equation:**
\[ R + h = R \cdot \sqrt{\frac{x}{y}} \]

8. **Solving for \( h \):**
\[ h = R \cdot \sqrt{\frac{x}{y}} - R \]
\[ h = R \left( \sqrt{\frac{x}{y}} - 1 \right) \]

Thus, the value of \( h \) is:
\[ h = R \left( \sqrt{\frac{x}{y}} - 1 \right) \]