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:
brainly.com/question/16705471
#SPJ4
We don't even need to know how many pulses were produced
in those 3 seconds.
The beginning of the first pulse took 3 seconds to travel
45 centimeters from the generator.
Its speed is (45 cm) / (3 sec) = 15 cm/sec.
Density = Mass divided by Volume
Here is your answer:
First find the notations:
2×10^-3
=0002
And...
2.5×10^4=25000
Then divide:
0002÷25000=8E-9
Your answer:
=8 x 10-8
Answer:
58.5 meters
Explanation:
1. Find your formula. Use distance = speed x time for this problem
2. Plug in the given information. d (for distance) = 9m/s^2 * 6.5 s
3. Multiply number AND units. d = 58.5m
4. Check to make sure units & numbers make sense. In this case check that the answer is a lot bigger than what we stated with and that our units go with distance
