Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.
<h3>What is the one-solar mass star?</h3>
A star having a mass equal to the mass of the Sun is called a one-solar mass star.
Its life track shows the luminous intensity as well as the surface temperature.
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The relationship between current and voltage and resistance is described by ohlm's law. This equation i=v/r tells that the current i flowing through a circuit is directly proportional to the voltage v, and inversely proportional to resistance r. This desceibes the relationship of voltage, current and resistance.
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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