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:
4.8mph
Explanation:
Speed= Distance/time
Speed= 26.2/5.5
= 4.76mph
( To the nearest tenth ) = 4.8mph
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Answer:
1.2 kg
Explanation:
Let UP ramp be the positive direction
F = ma
T - Wt || - Ff = m(0)
mg - Μgsinθ - μΜgcosθ = 0
m(9.8) - 13sin35 - 0.36(13)cos35 = 0
m = 13(sin35 + 0.36cos35) / 9.8
m = 1.15205... ≈ 1.2 kg
Answer:
2 m/s²
Explanation:
the equations of motion are
S= ut +½at²
v² = u²+ 2as
v = u + at
s = (u+v)/2 × t
From the parameters given
u = 0m/s this is because it starts from rest
Distance (s) = 9m
Time (t) = 3s
Based on this the first equation would be used
s = ut + ½at²
Input values
9 = 0×3 + ½ × a x 3²
9 = 0 + 9a/2
9 = 4.5a
Divide both sides by 4.5
a = 9 / 4.5 m/s²
a = 2 m/s²
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