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, we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity \( v_e \) from a planet is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 2: Relate mass and volume
The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
Since the mean density is the same as that of Earth, we can use this relationship.

### Step 3: Volume of the planet
Given that the volume of the planet is 27 times that of Earth:
\[ V_{planet} = 27 V_{Earth} \]

### Step 4: Express the escape velocity in terms of volume
Substituting the mass in the escape velocity formula:
\[ v_e = \sqrt{\frac{2G(\rho V)}{R}} \]

### Step 5: Relate radius and volume
For a sphere, the volume \( V \) is related to the radius \( R \) by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the radius can be expressed as:
\[ R = \left(\frac{3V}{4\pi}\right)^{1/3} \]

### Step 6: Find the relationship between escape velocities
We know that the escape velocity is proportional to \( \sqrt{\frac{V}{R}} \). Since the volume of the new planet is 27 times that of Earth, we can express the escape velocity of the new planet \( v_{e, planet} \) in terms of the escape velocity of Earth \( v_{e, Earth} \):
\[ \frac{v_{e, planet}}{v_{e, Earth}} = \sqrt{\frac{V_{planet}/R_{planet}}{V_{Earth}/R_{Earth}}} \]

### Step 7: Calculate the ratio of volumes and radii
Since the volume of the planet is 27 times that of Earth, we have:
\[ \frac{V_{planet}}{V_{Earth}} = 27 \]
The radius \( R \) is related to the volume as \( R \propto V^{1/3} \), so:
\[ \frac{R_{planet}}{R_{Earth}} = \left(\frac{V_{planet}}{V_{Earth}}\right)^{1/3} = 27^{1/3} = 3 \]

### Step 8: Substitute the ratios into the escape velocity equation
Now substituting back into the escape velocity ratio:
\[ \frac{v_{e, planet}}{v_{e, Earth}} = \sqrt{\frac{27}{1/3}} = \sqrt{27 \cdot 3} = \sqrt{81} = 9 \]

### Step 9: Calculate the escape velocity of the new planet
Given that the escape velocity from Earth \( v_{e, Earth} = 11.2 \, \text{km/s} \):
\[ v_{e, planet} = 3 \times v_{e, Earth} = 3 \times 11.2 \, \text{km/s} = 33.6 \, \text{km/s} \]

### Final Answer:
The escape speed from the planet is \( 33.6 \, \text{km/s} \).

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