Respuesta :
Answer:
Find the answer in the explanation.
Explanation:
Given that the international space station (ISS) is in a low earth orbit, just 400km above the surface. And at this altitude, the acceleration due to gravity has a value of 8.69.
A.) Since the radius of the earth is equal to 6400 km. At it to 400km
The radius of the orbit = 6400 + 400
The radius = 6800 km.
Where orbital speed = sqrt ( g × r )
Orbital speed = sqrt ( 6800 × 8.69 )
Orbital speed = 243.09 km/h
B.) Orbital period acceleration to Kepler third law of planetary motion state that:
The square of the period is directly proportional to the cube of the radius.
T^2 = (4π^2 /GM) r^3
T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3
T^2 = (4π^2/ 4.002^36) × 6800^3
T^2 = 9.865×10^-36 × 6800^3
T^2 = 3.10 × 10^-15
T = 5.57 × 10^-8 hours.
Convert the hours to minutes by multiplying it by 60
T = 3.4 × 10^-6 minutes.
Part A:
Since the radius of the earth is equal to 6400 km. At it to 400km
The radius of the orbit = 6400 + 400
The radius = 6800 km.
Where:
- Orbital speed = sqrt ( g × r )
- Orbital speed = sqrt ( 6800 × 8.69 )
- Orbital speed = 243.09 km/h
The speed of the ISS is 243.09Km/h.
Part B:
The orbital period (T) of the ISS in minutes is :
- T^2 = (4π^2 /GM) r^3
- T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3
- T^2 = (4π^2/ 4.002^36) × 6800^3
- T^2 = 9.865×10^-36 × 6800^3
- T^2 = 3.10 × 10^-15
- T = 5.57 × 10^-8 hours.
Convert the hours to minutes by multiplying it by 60
T = 3.4 × 10^-6 minutes.
The orbital period (T) of the ISS in minutes is 3.4 × 10^-6 minutes.
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