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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Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!