Suppose the earth was covered by an oceam of uniform depth `h,(hltltR)` Let `sigma` be densityof ocean and `p` be mean density of earth. Let `Deltag` be the approxemate difference of value of net acceleration duet to gravity between the bottom of the ocean and top `(Deltag=g_("top")-g_("bottom"))`. Choose the correct option.

Options

A. `Deltag=(4)/(3)piGh[2rho-3sigma]`
B. `Deltag=(4)/(3)Gpih[3sigma-2rho]`
C. `Deltag=(4)/(3)Gpih[2sigma-3rho]`
D. `Deltag=(4)/(3)Gpih[3rho-3sigma]`

Answer

To solve the problem, we need to find the difference in the acceleration due to gravity at the top of the ocean (g_top) and at the bottom of the ocean (g_bottom), denoted as Δg = g_top - g_bottom.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have an ocean of uniform depth \( h \) covering the Earth with radius \( R \).
- The density of the ocean is \( \sigma \) and the mean density of the Earth is \( p \).

2. **Acceleration Due to Gravity at the Bottom of the Ocean (g_bottom)**:
- The formula for acceleration due to gravity at a distance \( r \) from the center of the Earth is given by:
\[ g = \frac{GM}{r^2} \]
- At the bottom of the ocean, the distance from the center of the Earth is \( R \). Therefore:
\[ g_{bottom} = \frac{G M_{earth}}{R^2} \]
- The mass of the Earth \( M_{earth} \) can be expressed in terms of its density \( p \):
\[ M_{earth} = \frac{4}{3} \pi R^3 p \]
- Substituting this into the equation for \( g_{bottom} \):
\[ g_{bottom} = \frac{G \left(\frac{4}{3} \pi R^3 p\right)}{R^2} = \frac{4}{3} \pi G R p \]

3. **Acceleration Due to Gravity at the Top of the Ocean (g_top)**:
- At the top of the ocean, the distance from the center of the Earth is \( R + h \). Thus:
\[ g_{top} = \frac{G M_{total}}{(R + h)^2} \]
- The total mass \( M_{total} \) includes both the mass of the Earth and the mass of the ocean:
\[ M_{total} = M_{earth} + M_{ocean} \]
- The mass of the ocean can be calculated as:
\[ M_{ocean} = \sigma \cdot V_{ocean} = \sigma \cdot \left(\frac{4}{3} \pi (R + h)^3 - \frac{4}{3} \pi R^3\right) \]
- Simplifying this gives:
\[ M_{ocean} = \sigma \cdot \left(\frac{4}{3} \pi \left((R + h)^3 - R^3\right)\right) \]
- Therefore, substituting back into the equation for \( g_{top} \):
\[ g_{top} = \frac{G \left(\frac{4}{3} \pi R^3 p + \sigma \cdot \left(\frac{4}{3} \pi \left((R + h)^3 - R^3\right)\right)\right)}{(R + h)^2} \]

4. **Finding the Difference Δg**:
- Now we can find Δg:
\[ \Delta g = g_{top} - g_{bottom} \]
- Substituting the expressions we derived for \( g_{top} \) and \( g_{bottom} \):
\[ \Delta g = \frac{G \left(\frac{4}{3} \pi R^3 p + \sigma \cdot \left(\frac{4}{3} \pi \left((R + h)^3 - R^3\right)\right)\right)}{(R + h)^2} - \frac{4}{3} \pi G R p \]

5. **Simplifying the Expression**:
- After simplification, we can factor out common terms and apply the binomial approximation for small \( h \) compared to \( R \):
\[ \Delta g \approx \frac{4}{3} \pi G h \left(3\sigma - 2p\right) \frac{1}{R^2} \]

### Final Result:
The approximate difference in the value of net acceleration due to gravity between the bottom of the ocean and the top is:
\[ \Delta g \approx \frac{4}{3} \pi G h (3\sigma - 2p) \frac{1}{R^2} \]