Question

A micro meteor has a mass of 0.0055 G when it enters Earth's atmosphere and travels at 21,000 meters per second what is its kinetic energy when it enters Earth's atmosphere

Answer 1

To calculate the kinetic energy (KE) of the micro meteor when it enters Earth's atmosphere, we can use the kinetic energy formula:

\[
KE = \frac{1}{2}mv^2
\]

where:
- \( m \) is the mass of the meteor in kilograms,
- \( v \) is the velocity of the meteor in meters per second.

Given:
- The mass of the micro meteor \( m = 0.0055 \) grams \( = 0.0055 \times 10^{-3} \) kg \( = 5.5 \times 10^{-6} \) kg.
- The speed \( v = 21,000 \) m/s.

Now we can plug in the values into the kinetic energy equation:

\[
KE = \frac{1}{2} \times (5.5 \times 10^{-6} \, \text{kg}) \times (21,000 \, \text{m/s})^2
\]

Calculating \( v^2 \):

\[
v^2 = (21,000)^2 = 441,000,000 \, \text{m}^2/\text{s}^2
\]

Now substitute this back into the equation for kinetic energy:

\[
KE = \frac{1}{2} \times (5.5 \times 10^{-6}) \times 441,000,000
\]

Calculating this:

\[
KE = 0.5 \times 5.5 \times 441,000 \, \text{(since \( 10^{-6} \times 10^6 = 1 \))}
\]
\[
KE = 2.5 \times 441,000 = 1,102,500 \, \text{Joules}
\]

Therefore, the kinetic energy of the micro meteor when it enters Earth's atmosphere is approximately **1,102,500 Joules** or **1.1 MJ** (megajoule).