The acceleration due to gravity at the surface of the earth is g. Calculate its value at the surface of the sum. Given that the radius of sun is 110 times that of the earth and its mass is `33 xx 10^(4)` times that of the earth.

Explanation

To calculate the acceleration due to gravity at the surface of the Sun, we can use the formula for gravitational acceleration:

\[ g = \frac{G \cdot M}{R^2} \]

where:
- \( g \) is the acceleration due to gravity,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the object,
- \( R \) is the radius of the object.

### Step 1: Write the formula for the acceleration due to gravity at the surface of the Earth
The acceleration due to gravity at the surface of the Earth is given by:

\[ g_e = \frac{G \cdot M_e}{R_e^2} \]

where:
- \( g_e \) is the acceleration due to gravity at the surface of the Earth,
- \( M_e \) is the mass of the Earth,
- \( R_e \) is the radius of the Earth.

### Step 2: Write the formula for the acceleration due to gravity at the surface of the Sun
The acceleration due to gravity at the surface of the Sun can be expressed as:

\[ g_s = \frac{G \cdot M_s}{R_s^2} \]

where:
- \( g_s \) is the acceleration due to gravity at the surface of the Sun,
- \( M_s \) is the mass of the Sun,
- \( R_s \) is the radius of the Sun.

### Step 3: Substitute the values for mass and radius of the Sun
From the problem, we know that:
- The mass of the Sun \( M_s = 33 \times 10^4 \times M_e \)
- The radius of the Sun \( R_s = 110 \times R_e \)

### Step 4: Substitute these values into the formula for \( g_s \)
Now we can substitute \( M_s \) and \( R_s \) into the equation for \( g_s \):

\[ g_s = \frac{G \cdot (33 \times 10^4 \cdot M_e)}{(110 \cdot R_e)^2} \]

### Step 5: Simplify the equation
This simplifies to:

\[ g_s = \frac{G \cdot 33 \times 10^4 \cdot M_e}{110^2 \cdot R_e^2} \]

### Step 6: Relate \( g_s \) to \( g_e \)
We know that:

\[ g_e = \frac{G \cdot M_e}{R_e^2} \]

Thus, we can express \( g_s \) in terms of \( g_e \):

\[ g_s = g_e \cdot \frac{33 \times 10^4}{110^2} \]

### Step 7: Calculate the numerical value
Now we can calculate the value:

1. Calculate \( 110^2 = 12100 \).
2. Thus,

\[ g_s = g_e \cdot \frac{33 \times 10^4}{12100} \]

3. Since \( g_e \) is approximately \( 9.81 \, \text{m/s}^2 \):

\[ g_s = 9.81 \cdot \frac{33 \times 10^4}{12100} \]

4. Calculate \( \frac{33 \times 10^4}{12100} \):

\[ \frac{33 \times 10^4}{12100} \approx 273.3 \]

5. Therefore,

\[ g_s \approx 9.81 \cdot 273.3 \approx 267.3 \, \text{m/s}^2 \]

### Final Result
The acceleration due to gravity at the surface of the Sun is approximately:

\[ g_s \approx 27.3 \, g_e \]