Escape velocity at earth's surface is `11.2km//s`. Escape velocity at the surface of a planet having mass `100` times and radius `4` times that of earth will be

Options

A. `56km//s`
B. `112km//s`
C. `305km//s`
D. `11.2 km//s`

Answer

To find the escape velocity at the surface of a planet with mass \(100\) times and radius \(4\) times that of Earth, we can use the formula for escape velocity:

\[ v_e = \sqrt{\frac{2GM}{R}} \]

Where:
- \(v_e\) is the escape velocity,
- \(G\) is the gravitational constant,
- \(M\) is the mass of the planet,
- \(R\) is the radius of the planet.

### Step 1: Identify the escape velocity of Earth
Given that the escape velocity at Earth's surface is \(11.2 \, \text{km/s}\), we can express this as:

\[ v_{e, \text{Earth}} = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} = 11.2 \, \text{km/s} \]

### Step 2: Determine the mass and radius of the new planet
The new planet has:
- Mass \(M = 100 M_{\text{Earth}}\)
- Radius \(R = 4 R_{\text{Earth}}\)

### Step 3: Substitute the values into the escape velocity formula for the new planet
Now, we can calculate the escape velocity for the new planet:

\[ v_{e, \text{planet}} = \sqrt{\frac{2G(100 M_{\text{Earth}})}{4 R_{\text{Earth}}}} \]

### Step 4: Simplify the expression
This simplifies to:

\[ v_{e, \text{planet}} = \sqrt{\frac{100 \cdot 2GM_{\text{Earth}}}{4 R_{\text{Earth}}}} = \sqrt{\frac{100}{4} \cdot \frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} = \sqrt{25} \cdot \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} \]

### Step 5: Calculate the escape velocity
Since \(\sqrt{25} = 5\), we have:

\[ v_{e, \text{planet}} = 5 \cdot \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} = 5 \cdot v_{e, \text{Earth}} \]

Substituting \(v_{e, \text{Earth}} = 11.2 \, \text{km/s}\):

\[ v_{e, \text{planet}} = 5 \cdot 11.2 \, \text{km/s} = 56 \, \text{km/s} \]

### Final Answer
The escape velocity at the surface of the planet is \(56 \, \text{km/s}\).

---