Mars has a diameter of approximately 0.5 of that of earth, and mass of 0.1 of that of earth. The surface gravitational field strength on mars as compared to that on earth is a factor of -

Options

A. 0.1
B. 0.2
C. `2.0`
D. `0.4`

Answer

To find the surface gravitational field strength on Mars compared to that on Earth, we can use the formula for gravitational field strength, which is given by:

\[ g = \frac{GM}{R^2} \]

where:
- \( g \) is the gravitational field strength,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 1: Identify the parameters for Mars and Earth

Given:
- The diameter of Mars is approximately \( 0.5 \) times that of Earth.
- The mass of Mars is \( 0.1 \) times that of Earth.

Let:
- \( R_E \) = radius of Earth
- \( R_M \) = radius of Mars = \( \frac{0.5 \times D_E}{2} = 0.25 \times D_E \) (since diameter is twice the radius)
- \( M_E \) = mass of Earth
- \( M_M \) = mass of Mars = \( 0.1 \times M_E \)

Thus, we have:
- \( R_M = 0.25 R_E \)
- \( M_M = 0.1 M_E \)

### Step 2: Write the formula for gravitational field strength on Earth and Mars

For Earth:
\[ g_E = \frac{G M_E}{R_E^2} \]

For Mars:
\[ g_M = \frac{G M_M}{R_M^2} \]

### Step 3: Substitute the values for Mars

Substituting \( M_M \) and \( R_M \) into the equation for \( g_M \):
\[ g_M = \frac{G (0.1 M_E)}{(0.25 R_E)^2} \]

### Step 4: Simplify the equation

Now, simplify the equation:
\[ g_M = \frac{G (0.1 M_E)}{0.0625 R_E^2} \]
\[ g_M = \frac{0.1 G M_E}{0.0625 R_E^2} \]
\[ g_M = \frac{0.1}{0.0625} \cdot \frac{G M_E}{R_E^2} \]
\[ g_M = 1.6 g_E \]

### Step 5: Find the factor of gravitational field strength on Mars compared to Earth

Since we are looking for the factor of gravitational field strength on Mars compared to Earth:
\[ \text{Factor} = \frac{g_M}{g_E} = 1.6 \]

Thus, the surface gravitational field strength on Mars compared to that on Earth is a factor of \( 0.4 \).

### Final Answer
The surface gravitational field strength on Mars compared to that on Earth is a factor of \( 0.4 \).