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:

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

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: Define the variables for Earth and Mars
Let:
- \( M_E \) = mass of Earth
- \( R_E \) = radius of Earth
- \( M_M \) = mass of Mars = 0.1 \( M_E \)
- \( R_M \) = radius of Mars

Since the diameter of Mars is approximately 0.5 that of Earth, we have:
\[ R_M = 0.5 R_E \]

### Step 2: Write the gravitational field strength equations
The gravitational field strength on Earth (\( g_E \)) is given by:
\[ g_E = \frac{G \cdot M_E}{R_E^2} \]

The gravitational field strength on Mars (\( g_M \)) is given by:
\[ g_M = \frac{G \cdot M_M}{R_M^2} \]

### Step 3: Substitute the values for Mars
Substituting \( M_M \) and \( R_M \) into the equation for \( g_M \):
\[ g_M = \frac{G \cdot (0.1 M_E)}{(0.5 R_E)^2} \]

### Step 4: Simplify the equation for \( g_M \)
Now simplify the equation:
\[ g_M = \frac{G \cdot (0.1 M_E)}{0.25 R_E^2} \]
\[ g_M = \frac{0.1 G \cdot M_E}{0.25 R_E^2} \]
\[ g_M = \frac{0.1}{0.25} \cdot \frac{G \cdot M_E}{R_E^2} \]
\[ g_M = 0.4 \cdot g_E \]

### Step 5: Conclusion
Thus, the surface gravitational field strength on Mars compared to that on Earth is a factor of:
\[ g_M = 0.4 \cdot g_E \]

### Final Answer
The surface gravitational field strength on Mars is 0.4 times that on Earth.

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