The value of acceleration due to gravity on the surface of earth is x. At altitude 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, we need to find the height \( h \) at which the acceleration due to gravity \( y \) is measured, given that the acceleration due to gravity at the surface of the Earth is \( x \) and the radius of the Earth is \( R \).

### Step-by-Step Solution:

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

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

3. **Setting up the ratio of the two equations**:
To eliminate \( GM \), we can take the ratio of Equation 1 and Equation 2:
\[ \frac{x}{y} = \frac{GM/R^2}{GM/(R + h)^2} \]
This simplifies to:
\[ \frac{x}{y} = \frac{(R + h)^2}{R^2} \]

4. **Cross-multiplying and simplifying**:
Cross-multiplying gives:
\[ x(R^2) = y(R + h)^2 \]
Expanding the right side:
\[ xR^2 = y(R^2 + 2Rh + h^2) \]

5. **Rearranging the equation**:
Rearranging gives:
\[ xR^2 = yR^2 + 2yRh + yh^2 \]
Bringing all terms involving \( h \) to one side:
\[ 0 = (y - x)R^2 + 2yRh + yh^2 \]

6. **Solving for \( h \)**:
This is a quadratic equation in \( h \):
\[ yh^2 + 2yRh + (y - x)R^2 = 0 \]
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = y \), \( b = 2yR \), and \( c = (y - x)R^2 \):
\[ h = \frac{-2yR \pm \sqrt{(2yR)^2 - 4y(y - x)R^2}}{2y} \]
Simplifying this gives:
\[ h = \frac{-2yR \pm 2R\sqrt{y}}{2y} \]
\[ h = R\left(\sqrt{\frac{x}{y}} - 1\right) \]

7. **Final expression for \( h \)**:
Thus, the height \( h \) is given by:
\[ h = R\left(\sqrt{\frac{x}{y}} - 1\right) \]

### Final Answer:
\[ h = R\left(\sqrt{\frac{x}{y}} - 1\right) \]