A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass `=5.98xx10^(24)kg`) have to be compresed to be a black hole?

Options

A. `10^(-9)m`
B. `10^(-6)m`
C. `10^(-2)m`
D. `100m`

Answer

To determine the radius to which Earth must be compressed to become a black hole, we need to find the Schwarzschild radius (the radius of a black hole) using the escape velocity formula. Here’s the step-by-step solution:

### Step 1: Understand the concept of escape velocity
The escape velocity \( v \) from the surface of a mass \( M \) at a distance \( r \) is given by the formula:
\[ v = \sqrt{\frac{2GM}{r}} \]
where \( G \) is the gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2 \).

### Step 2: Set the escape velocity equal to the speed of light
For Earth to become a black hole, the escape velocity must equal the speed of light \( c \):
\[ c = 3 \times 10^8 \, \text{m/s} \]
Thus, we set:
\[ 3 \times 10^8 = \sqrt{\frac{2GM}{r}} \]

### Step 3: Square both sides of the equation
Squaring both sides to eliminate the square root gives:
\[ (3 \times 10^8)^2 = \frac{2GM}{r} \]
Calculating \( (3 \times 10^8)^2 \):
\[ 9 \times 10^{16} = \frac{2GM}{r} \]

### Step 4: Rearrange the equation to solve for \( r \)
Rearranging the equation to isolate \( r \):
\[ r = \frac{2GM}{9 \times 10^{16}} \]

### Step 5: Substitute the known values
Substituting \( G = 6.67 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2 \) and \( M = 5.98 \times 10^{24} \, \text{kg} \):
\[ r = \frac{2 \times (6.67 \times 10^{-11}) \times (5.98 \times 10^{24})}{9 \times 10^{16}} \]

### Step 6: Calculate the numerator
Calculating the numerator:
\[ 2 \times 6.67 \times 5.98 \approx 79.76 \quad \text{(approximately)} \]
Thus, the numerator becomes:
\[ 79.76 \times 10^{13} \quad \text{(since } 10^{-11} \times 10^{24} = 10^{13}\text{)} \]

### Step 7: Calculate \( r \)
Now substituting back:
\[ r = \frac{79.76 \times 10^{13}}{9 \times 10^{16}} = \frac{79.76}{9} \times 10^{-3} \]
Calculating \( \frac{79.76}{9} \):
\[ \approx 8.86 \quad \text{(approximately)} \]
Thus:
\[ r \approx 8.86 \times 10^{-3} \, \text{m} \]

### Step 8: Final approximation
The approximate radius to which Earth must be compressed to become a black hole is:
\[ r \approx 8.86 \times 10^{-3} \, \text{m} \approx 10^{-2} \, \text{m} \]

### Conclusion
Therefore, the answer is approximately \( 10^{-2} \, \text{m} \).

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