To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,
![F_g = F_c](https://tex.z-dn.net/?f=F_g%20%3D%20F_c)
![\frac{GmM}{r^2} = \frac{mv^2}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BGmM%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D)
Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity
![\frac{GM}{r^2} = \frac{v^2}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BGM%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D)
![\frac{GM}{r} = v^2](https://tex.z-dn.net/?f=%5Cfrac%7BGM%7D%7Br%7D%20%3D%20v%5E2)
![v = \sqrt{\frac{GM}{r}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7Br%7D%7D)
PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,
![v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B%286.67%2A10%5E%7B-11%7D%29%285.97%2A10%5E%7B24%7D%29%7D%7B3%2A%286.371%2A10%5E6%29%7D%7D)
![v = 4564.42m/s](https://tex.z-dn.net/?f=v%20%3D%204564.42m%2Fs)
From the speed it is possible to use find the formula, so
![T = \frac{2\pi r}{v}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20r%7D%7Bv%7D)
![T = \frac{2\pi (6.371*10^6)}{4564.42}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20%286.371%2A10%5E6%29%7D%7B4564.42%7D)
![T = 8770.05s\approx 146min\approx 2.4hour](https://tex.z-dn.net/?f=T%20%3D%208770.05s%5Capprox%20146min%5Capprox%202.4hour)
Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.
PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is
![KE = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
![KE = \frac{1}{2} (100)(4564.42)^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28100%29%284564.42%29%5E2)
![KE = 1.0416*10^9J](https://tex.z-dn.net/?f=KE%20%3D%201.0416%2A10%5E9J)
Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
![V^2 = U^2 + 2dS](https://tex.z-dn.net/?f=V%5E2%20%3D%20U%5E2%20%2B%202dS)
Substituting the values into the formula, we have;
![V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}](https://tex.z-dn.net/?f=V%5E2%20%3D%2022%20%2B%202%281.4%29%28110%29%5C%5C%5C%5CV%5E2%20%3D%2022%20%2B%20308%5C%5C%5C%5CV%5E2%20%3D%20330%5C%5C%5C%5CV%5E2%20%3D%20%5Csqrt%7B330%7D)
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
Frequensey or hertz, I looked this up on the internet!
Answer:
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