Respuesta :
Answer:
The orbital speed is approximately 17,325.57 m/s
The number of Earth days it would take lo to complete its orbit is approximately 1.77 days
Explanation:
The given parameters are;
The mass of lo, m ≈ 7.22 × 10²² kg
The radius of lo, R ≈ 1.82 × 10⁶ m
The mean distance between Jupiter and lo = 4.22 × 10⁸ m
The orbital equation is given as follows;
[tex]\dfrac{m \cdot v^2}{R} = \dfrac{G \times M \times m}{R^2}[/tex]
[tex]\therefore v = \sqrt{\dfrac{G \cdot M}{R} } = \sqrt{\dfrac{ 6.67408 \times 10^{-11} \times 1.898 \times 10^{27}}{4.22 \times 10^8} } = 17,325.57 \ m/s[/tex]
The orbital speed ≈ 17,325.57 m/s
The time to complete one orbit = (2 × π × 4.22 × 10⁸)/(17325.57) ≈ 153039.94 s
The time to complete one orbit ≈ 153039.94 s ≈ 1.77 days
The number of Earth days it would take lo to complete its orbit ≈ 1.77 days.
The number of Earth days it would take lo to complete its orbit will be 1.77 days. While the orbital speed will be 17,325.57 m/s.
What is orbital speed?
The speed required to establish an equilibrium between gravity's pull on the satellite and the inertia of the satellite's motion is referred to as orbital speed.
The given data in the problem is;
m is the mass of lo≈ 7.22 × 10²² kg
R is the radius of lo ≈ 1.82 × 10⁶ m
The mean distance between Jupiter and lo = 4.22 × 10⁸ m
The orbital speed is given as;
[tex]\rm V= \sqrt{\frac{GM}{R} } \\\\ \rm V= \sqrt{\frac{6.67 \times 10^{-11} \times 1.898 \times 10^{27}}{4.22 \times 10^8} } \\\\ \rm V=17,325.57 \ m/sec[/tex]
The time to complete one orbit is given as;
[tex]\rm T=\frac{ (2 \times \pi \times 4.22 \times 10^8)}{17325.57 } \\\\ \rm T= 153039.94 \ s[/tex]
In one day there are 86400 seconds from that calculation the number of days required to complete one orbit will be 1.77 days
Hence the number of Earth days it would take lo to complete its orbit will be 1.77 days. While the orbital speed will be 17,325.57 m/s.
To learn more about the orbital speed refer to the link;
https://brainly.com/question/541239
