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.
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Explanation:
t = usin©/g
Where t is the time to reach the maximum height
Time spent in air is T = 2t
Hence, T = 2usin©/g
T = 2 x 20 x sin 65°/ 9.8
T = 3.69s
The amount of extra electrons present on the negative surface is
57.2 x 
.
Distance =1.6 cm
Side = 24 cm
Electric field = 18000 N/C
Calculating the capacitance in the metal plates is necessary.
Using the capacitance formula
Putting the value
C = 8.85 x 
x (24 x 
/1.6 x
C = 0.318 x 
F
<h3>Calculation of potential</h3>
V = Ed
V = 18000 x 1.6 x V
V = 288 V
<h3>Calculation of charge</h3>
Q = CV
Q = 0.318 x x 288
Q = 91.54 x C
Charge on the both the plates
Q = +91.54 x
Q = - 91.54 x
Calculation of excess electrons on the negative surface:
n = q/e
n = 91.54 x / 1.6 x 
n = 57.2 x electrons
Hence, the number of excess electrons on the negative surface is
57.2 x .
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