It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.
To find the answer, we need to know about the third law of Kepler.
<h3>What's the Kepler's third law?</h3>
- It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
- Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
- The time period of geosynchronous orbit is 24 hours or 1440 minutes.
- As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
- If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
- a1= (T1/T2)⅔×a2
= (1440/90)⅔×6780
= 43,090 km
- Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km
Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.
Learn more about the Kepler's third law here:
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1200 watt seconds
1.2. Kw seconds
1.2/ 3600 KWh
C it takes a lot of capital to get the windmills and the fields of windmills set up
Answer:
Acceleration = (change in speed) / (time for the change)
Change in speed = (ending speed) - (beginning speed)
= (3,600 mi/hr) - ( 0 )
= 3,600 mi/hr
= same as 1 mile/second.
Acceleration = (1 mi/sec) / (10 sec)
= 0.1 mi/sec² .
But 1 mile = 5,280 ft,
so
0.1 mi/s² = 528 ft/s²
Explanation:
have a great day
Answer:
The product of (wavelength) x (frequency) is always the same number ... the wave's speed.
So if the wavelength is somehow reduced to 1/4 its original length, the frequency is immediately multiplied by 4 . That's the only way their product can remain the same.
Explanation:
