The value of acceleration due to gravity on the surface of earth is x. At alttitude of .h. from the surface of 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`

Answer

To solve the problem of finding the height \( h \) at which the acceleration due to gravity changes from \( x \) on the surface of the Earth to \( y \) at height \( h \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formulas**:
The acceleration due to gravity at the surface of the Earth is given by:
\[ x = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Acceleration at Height \( h \)**:
The acceleration due to gravity at a height \( h \) above the surface of the Earth is given by:
\[ y = \frac{GM}{(R + h)^2} \]

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

4. **Dividing the Two Equations**:
To eliminate \( GM \), we can divide Equation 1 by Equation 2:
\[ \frac{x}{y} = \frac{(R + h)^2}{R^2} \]

5. **Cross-Multiplying**:
Rearranging gives:
\[ x(R^2) = y(R + h)^2 \]

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

7. **Rearranging the Equation**:
This can be rearranged to:
\[ R + h = R\sqrt{\frac{x}{y}} \]

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

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