What should be the radius of a planet with mass equal to that of earth and escape velocity on its surface is equal to the velocity of light. Given that mass of earth is `M=6xx10^(24)kg`.

Options

A. 6mm
B. 9mm
C. 18mm
D. 3mm

Explanation

To find the radius of a planet with a mass equal to that of Earth and an escape velocity equal to the speed of light, we can follow these steps:

### Step 1: Understand the formula for escape velocity
The escape velocity \( V_e \) from the surface of a planet is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the gravitational constant \( (6.67 \times 10^{-11} \, \text{m}^3/\text{kg s}^2) \)
- \( M \) is the mass of the planet
- \( R \) is the radius of the planet

### Step 2: Set the escape velocity equal to the speed of light
According to the problem, the escape velocity is equal to the speed of light \( c \):
\[ c = 3 \times 10^8 \, \text{m/s} \]
Thus, we can set:
\[ \sqrt{\frac{2GM}{R}} = c \]

### Step 3: Rearrange the formula to solve for radius \( R \)
Squaring both sides gives:
\[ \frac{2GM}{R} = c^2 \]
Rearranging for \( R \):
\[ R = \frac{2GM}{c^2} \]

### Step 4: Substitute known values
We know:
- \( M = 6 \times 10^{24} \, \text{kg} \) (mass of Earth)
- \( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Substituting these values into the equation for \( R \):
\[ R = \frac{2 \times (6.67 \times 10^{-11}) \times (6 \times 10^{24})}{(3 \times 10^8)^2} \]

### Step 5: Calculate the numerator and denominator
Calculating the numerator:
\[ 2 \times (6.67 \times 10^{-11}) \times (6 \times 10^{24}) = 8.004 \times 10^{14} \]

Calculating the denominator:
\[ (3 \times 10^8)^2 = 9 \times 10^{16} \]

### Step 6: Calculate \( R \)
Now, substituting back into the equation:
\[ R = \frac{8.004 \times 10^{14}}{9 \times 10^{16}} = 0.08893 \, \text{m} \approx 0.089 \, \text{m} \approx 89 \, \text{cm} \]

### Conclusion
Thus, the radius of the planet is approximately \( 89 \, \text{cm} \).