The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

Options

A. `9.8 m//sec^(2)`
B. `4.9 m//sec^(2)`
C. `980 m//sec^(2)`
D. `19.6 m//sec^(2)`

Answer

To find the acceleration due to gravity on the surface of a planet with twice the mass and diameter of Earth, we can follow these steps:

### Step 1: Understand the given parameters
- Let the mass of Earth be \( m \).
- The mass of the planet is given as \( 2m \).
- Let the radius of Earth be \( r \).
- The radius of the planet is given as \( 2r \) (since diameter is twice, radius is also doubled).

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

### Step 3: Substitute the values for the planet
For the planet, we substitute \( M = 2m \) and \( R = 2r \):
\[ g' = \frac{G(2m)}{(2r)^2} \]

### Step 4: Simplify the expression
Now simplify the equation:
\[ g' = \frac{2Gm}{4r^2} = \frac{Gm}{2r^2} \]

### Step 5: Relate it to Earth's gravity
We know that the acceleration due to gravity on Earth \( g = \frac{GM}{R^2} = \frac{Gm}{r^2} \). Therefore, we can express \( g' \) in terms of \( g \):
\[ g' = \frac{1}{2} \cdot \frac{Gm}{r^2} = \frac{g}{2} \]

### Step 6: Substitute the value of Earth's gravity
Given that \( g \) (acceleration due to gravity on Earth) is approximately \( 9.8 \, \text{m/s}^2 \):
\[ g' = \frac{9.8}{2} = 4.9 \, \text{m/s}^2 \]

### Conclusion
The acceleration due to gravity on the surface of the planet is \( 4.9 \, \text{m/s}^2 \).

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