What will be the escape speed from, a planet having volume 27 times that of earth and the same mean density as that of the earth? Take escape speed from the earth 11.2 km/s, earth and planet are perfectly sphere)

Options

A. 5.6 km/s
B. 22.4 km/s
C. 11.2 km/s
D. 33.6 km/s

Answer

To find the escape speed from a planet with a volume 27 times that of Earth and the same mean density as Earth, we can follow these steps:

### Step 1: Understand the relationship between volume, radius, and density
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 2: Relate the volumes of the planet and Earth
Let \( V_e \) be the volume of Earth and \( V_p \) be the volume of the planet. According to the problem:
\[ V_p = 27 V_e \]
This means:
\[ \frac{4}{3} \pi r_p^3 = 27 \left( \frac{4}{3} \pi r_e^3 \right) \]
where \( r_e \) is the radius of Earth and \( r_p \) is the radius of the planet.

### Step 3: Simplify the volume equation
Cancelling \( \frac{4}{3} \pi \) from both sides gives:
\[ r_p^3 = 27 r_e^3 \]
Taking the cube root of both sides, we find:
\[ r_p = 3 r_e \]

### Step 4: Calculate the mass of the planet
Since the density of the planet is the same as that of Earth, we can express the mass of the planet \( m_p \) in terms of the mass of Earth \( m_e \):
\[ \text{Density} = \frac{m}{V} \]
Thus, for the planet:
\[ \text{Density} = \frac{m_p}{V_p} = \frac{m_e}{V_e} \]
From the volume relationship:
\[ m_p = \text{Density} \times V_p = \text{Density} \times 27 V_e = 27 m_e \]

### Step 5: Use the escape velocity formula
The escape velocity \( v_e \) from a celestial body is given by:
\[ v_e = \sqrt{\frac{2 G m}{r}} \]
For Earth, we know:
\[ v_{e, \text{Earth}} = \sqrt{\frac{2 G m_e}{r_e}} = 11.2 \text{ km/s} \]

### Step 6: Substitute the mass and radius of the planet into the escape velocity formula
For the planet, we substitute \( m_p \) and \( r_p \):
\[ v_{e, \text{planet}} = \sqrt{\frac{2 G (27 m_e)}{3 r_e}} = \sqrt{27} \cdot \sqrt{\frac{2 G m_e}{r_e}} \cdot \frac{1}{\sqrt{3}} \]
This simplifies to:
\[ v_{e, \text{planet}} = \sqrt{9} \cdot v_{e, \text{Earth}} = 3 v_{e, \text{Earth}} \]

### Step 7: Calculate the escape velocity for the planet
Substituting the value of \( v_{e, \text{Earth}} \):
\[ v_{e, \text{planet}} = 3 \times 11.2 \text{ km/s} = 33.6 \text{ km/s} \]

### Final Answer
The escape speed from the planet is:
\[ \boxed{33.6 \text{ km/s}} \]

---