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 follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The acceleration due to gravity \( g \) at the surface of a planet is given by the formula:
\[ g = \frac{G M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 2: Define the parameters for Earth
For Earth, we have:
- Mass of Earth = \( M_E \)
- Radius of Earth = \( R_E \)
- Acceleration due to gravity on Earth = \( g_E = 9.8 \, \text{m/s}^2 \)

### Step 3: Define the parameters for the new planet
According to the problem:
- Mass of the planet \( M = 3 M_E \)
- Diameter of the planet = 3 times the diameter of Earth, which means:
\[ \text{Diameter of the planet} = 3 D_E \implies \text{Radius of the planet} = \frac{3 D_E}{2} = 3 R_E \]

### Step 4: Substitute the values into the formula for the new planet
Now we can substitute these values into the formula for acceleration due to gravity:
\[ g = \frac{G (3 M_E)}{(3 R_E)^2} \]

### Step 5: Simplify the equation
Now simplify the equation:
\[ g = \frac{3 G M_E}{9 R_E^2} \]
This simplifies to:
\[ g = \frac{1}{3} \cdot \frac{G M_E}{R_E^2} \]

### Step 6: Relate to Earth's gravity
From the formula for Earth's gravity, we know:
\[ \frac{G M_E}{R_E^2} = g_E \]
Thus, we can write:
\[ g = \frac{1}{3} g_E \]

### Step 7: Calculate the value of \( g \)
Substituting the value of \( g_E \):
\[ g = \frac{1}{3} \times 9.8 \, \text{m/s}^2 = 3.2667 \, \text{m/s}^2 \]

### Step 8: Round the answer
Rounding this to one decimal place gives:
\[ g \approx 3.3 \, \text{m/s}^2 \]

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

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