The mass of a planet and its diameter are three times those of earth's. Then the acceleration due to gravity on the surface of the planet is : `(g =9.8 ms^(-2))`

Options

A. `3.3 ms^(-2)`
B. `4.9ms^(-2)`
C. `19.6ms^(-2)`
D. `29.4ms^(-2)`

Answer

To find the acceleration due to gravity on the surface of a planet whose mass and diameter are three times those of Earth, we can use the formula for gravitational acceleration:

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

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

### Step 1: Identify the mass and radius of the planet
Given that the mass of the planet \( M' \) is three times the mass of Earth \( M \):
\[ M' = 3M \]

The diameter of the planet is also three times that of Earth. Since the radius is half of the diameter, we have:
\[ R' = \frac{3 \times \text{Diameter of Earth}}{2} = 3R \]

### Step 2: Substitute the values into the formula
Now, substituting \( M' \) and \( R' \) into the formula for \( g' \):
\[ g' = \frac{G \cdot (3M)}{(3R)^2} \]

### Step 3: Simplify the equation
Now, simplify the equation:
\[ g' = \frac{G \cdot (3M)}{9R^2} \]
\[ g' = \frac{3G \cdot M}{9R^2} \]
\[ g' = \frac{1}{3} \cdot \frac{G \cdot M}{R^2} \]

### Step 4: Relate it to Earth's gravity
We know that the acceleration due to gravity on Earth \( g \) is given by:
\[ g = \frac{G \cdot M}{R^2} \]
Thus, we can express \( g' \) in terms of \( g \):
\[ g' = \frac{1}{3} g \]

### Step 5: Calculate \( g' \)
Now substituting the value of \( g \):
\[ g' = \frac{1}{3} \cdot 9.8 \, \text{m/s}^2 \]
\[ g' = 3.27 \, \text{m/s}^2 \]

### Conclusion
The acceleration due to gravity on the surface of the planet is approximately \( 3.3 \, \text{m/s}^2 \).