To determine the amount of kinetic energy possessed by Ceres, we will use the formula for kinetic energy (KE):
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where:
- \( m \) is the mass of the object
- \( v \) is the velocity of the object
Given the data:
- The mass of Ceres \( m = 3.0 \times 10^{21} \) kg
- The velocity of Ceres \( v = 17900 \) m/s
Let's plug these values into the formula and perform the calculation step-by-step:
1. Square the velocity (\( v^2 \)):
[tex]\[
v^2 = (17900 \, \text{m/s})^2 = 320410000 \, (\text{m/s})^2
\][/tex]
2. Multiply the mass (\( m \)) by the squared velocity (\( v^2 \)):
[tex]\[
m \times v^2 = (3.0 \times 10^{21} \, \text{kg}) \times (320410000 \, (\text{m/s})^2)
\][/tex]
When you do this multiplication, you get:
[tex]\[
3.0 \times 10^{21} \times 320410000 = 9.6123 \times 10^{29} \, \text{kg} \cdot (\text{m/s})^2
\][/tex]
3. Multiply the result by \( \frac{1}{2} \) to find the kinetic energy:
[tex]\[
KE = \frac{1}{2} \times 9.6123 \times 10^{29} \, \text{kg} \cdot (\text{m/s})^2 = 4.80615 \times 10^{29} \, \text{Joules}
\][/tex]
Thus, the amount of kinetic energy possessed by Ceres is:
[tex]\[ 4.80615 \times 10^{29} \, \text{Joules} \][/tex]
