To calculate the kinetic energy (KE) of the micrometeor, we can use the formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m \) is the mass in kilograms
- \( v \) is the velocity in meters per second
First, we need to convert the mass from grams to kilograms:
\[ m = 0.005 , \text{grams} = 0.005 , \text{grams} \times \frac{1 , \text{kg}}{1000 , \text{grams}} = 0.000005 , \text{kg} \]
Now, substitute the values into the equation:
\[ v = 21,000 , \text{m/s} \]
\[ KE = \frac{1}{2} (0.000005 , \text{kg}) (21,000 , \text{m/s})^2 \]
Calculating \( v^2 \):
\[ (21,000 , \text{m/s})^2 = 441,000,000 , \text{m}^2/\text{s}^2 \]
Now substituting this back into the equation for KE:
\[ KE = \frac{1}{2} (0.000005) (441,000,000) \] \[ KE = 0.0000025 \times 441,000,000 \] \[ KE = 1,102.5 , \text{J} \]
So the kinetic energy of the micrometeor when it enters Earth's atmosphere is:
\[ \text{Answer: } 1,102.5 , \text{J} \]