Radius of earth is 6400 km and that of mars is 3200 km. Mass of mars is 0.1 that of earth's mass. Then the acceleration due to gravity on mars is nearly

Options

A. `1 m//s^(2)`
B. `2.5 m//s^(2)`
C. `4 m//s^(2)`
D. `5 m//s^(2)`

Explanation

To find the acceleration due to gravity on Mars, we can use the formula for gravitational acceleration on the surface of a planet:

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

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

### Step 1: Identify the known values
- Radius of Earth, \( R_E = 6400 \) km = \( 6400 \times 10^3 \) m
- Radius of Mars, \( R_M = 3200 \) km = \( 3200 \times 10^3 \) m
- Mass of Mars, \( M_M = 0.1 M_E \) (where \( M_E \) is the mass of Earth)

### Step 2: Write the formula for gravity on Mars
Using the formula for gravitational acceleration, we can express the acceleration due to gravity on Mars as:

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

### Step 3: Substitute the mass of Mars
Substituting \( M_M = 0.1 M_E \) into the equation:

\[ g_M = \frac{G (0.1 M_E)}{R_M^2} \]

### Step 4: Substitute the radius of Mars
Now substituting \( R_M = 3200 \times 10^3 \) m into the equation:

\[ g_M = \frac{G (0.1 M_E)}{(3200 \times 10^3)^2} \]

### Step 5: Relate Mars' gravity to Earth's gravity
We know the acceleration due to gravity on Earth is:

\[ g_E = \frac{G M_E}{R_E^2} \approx 10 \, \text{m/s}^2 \]

Now we can express \( g_M \) in terms of \( g_E \):

\[ g_M = 0.1 \cdot \frac{G M_E}{(3200 \times 10^3)^2} = 0.1 \cdot g_E \cdot \frac{R_E^2}{R_M^2} \]

### Step 6: Calculate the ratio of the radii
Now we calculate the ratio of the radii:

\[ \frac{R_E}{R_M} = \frac{6400 \times 10^3}{3200 \times 10^3} = 2 \]

Thus,

\[ \frac{R_E^2}{R_M^2} = 2^2 = 4 \]

### Step 7: Substitute back into the equation
Now substituting this back into the equation for \( g_M \):

\[ g_M = 0.1 \cdot g_E \cdot 4 \]

### Step 8: Substitute the value of \( g_E \)
Substituting \( g_E \approx 10 \, \text{m/s}^2 \):

\[ g_M = 0.1 \cdot 10 \cdot 4 = 4 \, \text{m/s}^2 \]

### Conclusion
The acceleration due to gravity on Mars is nearly \( 4 \, \text{m/s}^2 \).

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