If `R_(E )` be the radius of Earth ,then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given `:rltR_( E))`

Options

A. `1-(r)/(R_(E))-(r^(2))/(R_(E)^(2))-(r^(3))/(R_(E)^(3))`
B. `1+(r)/(R_(E))-(r^(2))/(R_(E)^(2))-(r^(3))/(R_(E)^(3))`
C. `1+(r)/(R_(E))-(r^(2))/(R_(E)^(2))+(r^(3))/(R_(E)^(3))`
D. `1+(r)/(R_(E))+(r^(2))/(R_(E)^(2))+(r^(3))/(R_(E)^(3))`

Answer

To solve the problem, we need to find the ratio of the acceleration due to gravity at a depth 'r' below the Earth's surface to that at a height 'r' above the Earth's surface.

### Step-by-Step Solution:

1. **Understanding the acceleration due to gravity at a height (g_a)**:
The formula for the acceleration due to gravity at a height 'h' above the Earth's surface is given by:
\[ g_a = \frac{GM}{(R_E + h)^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R_E \) is the radius of the Earth. For our case, \( h = r \):
\[ g_a = \frac{GM}{(R_E + r)^2} \]

2. **Understanding the acceleration due to gravity at a depth (g_d)**:
The formula for the acceleration due to gravity at a depth 'd' below the Earth's surface is given by:
\[ g_d = \frac{GM}{R_E^2} \left(1 - \frac{d}{R_E}\right) \]
For our case, \( d = r \):
\[ g_d = \frac{GM}{R_E^2} \left(1 - \frac{r}{R_E}\right) \]

3. **Finding the ratio of g_d to g_a**:
We need to find the ratio \( \frac{g_d}{g_a} \):
\[ \frac{g_d}{g_a} = \frac{\frac{GM}{R_E^2} \left(1 - \frac{r}{R_E}\right)}{\frac{GM}{(R_E + r)^2}} \]
Simplifying this, we can cancel \( GM \):
\[ \frac{g_d}{g_a} = \frac{(R_E + r)^2}{R_E^2} \left(1 - \frac{r}{R_E}\right) \]

4. **Expanding the ratio**:
Now, we can expand the expression:
\[ \frac{g_d}{g_a} = \left(\frac{(R_E + r)^2}{R_E^2}\right) \left(1 - \frac{r}{R_E}\right) \]
\[ = \left(\frac{R_E^2 + 2R_Er + r^2}{R_E^2}\right) \left(1 - \frac{r}{R_E}\right) \]

5. **Further simplification**:
Now, we can simplify this expression:
\[ = \left(1 + \frac{2r}{R_E} + \frac{r^2}{R_E^2}\right) \left(1 - \frac{r}{R_E}\right) \]
\[ = 1 + \frac{2r}{R_E} + \frac{r^2}{R_E^2} - \frac{r}{R_E} - \frac{2r^2}{R_E^2} - \frac{r^3}{R_E^3} \]
\[ = 1 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3} \]

6. **Final Result**:
Therefore, the final ratio of the acceleration due to gravity at depth 'r' below the Earth's surface to that at height 'r' above the Earth's surface is:
\[ \frac{g_d}{g_a} = \frac{1 - \frac{r}{R_E}}{1 + \frac{r}{R_E}} \]