Respuesta :
Answer:
see explanation
Explanation:
In order to do this exercise, we need to use the expression for conservation of energy. In this case:
Total Energy = Potential Energy + Kinetic energy
The kinetic energy: 1/2 mV²
potential energy: GMm/R
As the meteor does impact on earth, the total energy at the end is 0, so:
1/2mV² = GMm/R
So, we have the speed of meteor 1 and we want the speed of impact, so the above expression can be rewritten for:
1/2m(V2² - V1²) = GMm (1/R - 1/r)
Where:
R: Radius of Earth
r: distance of the moon
M: Mass of earth
m: mass of meteroid
G: gravitational constant
With these data, and the fact that the speed of the meteor is 2 km/s (2000 m/s), we can solve for V2:
1/2m(V2² - V1²) = GMm (1/R - 1/r) ----> simplifying m we have:
1/2(V2² - V1²) = GM(1/R - 1/r)
V2² - V1² = 2GM (1/R - 1/r)
V2² = 2GM(1/R - 1/r) + V1²
Replacing all the values we have:
V2² = 2 * 6.67x10^-11 * 6x10^24(1/6.37x10^6 - 1/3.84x10^8)
V2² = 8x10^14 (1.54x10^-7) + 2000²
V2 = 11278.3 m/s or 11.28 km/s
b) in this part, we use the same expression from above, but we have to add the distance of the meteor that misses the earth, in this case 5000 km or 5x10^6 m. This is added to the radius of earth. Thus, we have:
V2² = 2 * 6.67x10^-11 * 6x10^24(1/6.37x10^6 + 5x10^6 - 1/3.84x10^8) + 2000²
V2²= 8x10^14(8.54x10^-8) + 2000²
V2 = 8.5 x 10^3 km/s
