Astrophysics- Calculate the altitude of a geosynchronous orbit

In summary, to calculate the altitude of a geosynchronous orbit with an orbital period of one sidereal day, you can use the equation a= (mu*(1/(2pi))^2)^(1/3). This will give you an altitude of approximately 21.6, assuming a circular orbit. The altitude of this geosynchronous satellite is significantly higher than the average altitude of the International Space Station, which is around 354 km. To calculate the orbital period of the International Space Station, you can use the equation T = ((2\pi)/(mu)a^(3/2))^(1/3), where mu is the mass of the Earth. Be careful to use the correct units in these calculations.
  • #1
sleepyhead212
2
0
a)Calculate the altitude of a geosynchronous orbit (an orbit that has an orbital period of one sidereal day)(altitude is measured from the surface of the earth/ The "r" in gravitational equations is always measured from the center of mass of an object. one may assume a circular orbit.)

b) how does the altitude of this geosynchronous satellite compare to the average 354 km altitude of the International Space Station?

c) Calculate the orbital period of the international Space Station (assume circular Orbit)



this was my attempt

a= (mu*(1/(2pi))^2)^(1/3)
a= (398600(1/(2pi))^2)^(1/3)
a=21.6

but for b) I am not sure if my prof. wants a word explanation or to solve the actual difference so i provided a word explination

then for c)

T = ((2\pi)/(mu)a^(3/2))^(1/3)
but am unsure what mu would be.

i'm not sure if what I'm doing is correct because we never went over this in class so i tried do this based off internet research. any help would be appreciated
 
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  • #2
The equation you have is Keplers third law (ignoring the mass of the satelite)
( period / 2pi )^2 = radius^3 / GM
If you plugin G and M=mass of the Earth you should get the right answer, be careful of the units.
 
  • #3
can this equation be used for both part a and c?
 
  • #4
The law describing their behaviour is the same, you simple have to find radius in one case given the period and the period in the other case given the radius.
 

1. What is a geosynchronous orbit?

A geosynchronous orbit is a type of orbit around a celestial body - typically a planet or moon - in which the orbiting object rotates at the same speed as the object's rotation, resulting in the orbiting object remaining in the same position relative to the surface of the body.

2. How is the altitude of a geosynchronous orbit calculated?

The altitude of a geosynchronous orbit can be calculated using the following formula: altitude = (radius of the celestial body + desired distance from the surface) * (orbital speed of the body / orbital speed of the object).

3. What is the desired distance from the surface in the calculation?

The desired distance from the surface is typically equal to the radius of the celestial body. For example, if calculating the altitude of a geosynchronous orbit around Earth, the desired distance would be 6,371 kilometers (radius of Earth).

4. What is the orbital speed of the body in the calculation?

The orbital speed of the body refers to the speed at which the celestial body rotates on its axis. In the case of Earth, the orbital speed is approximately 1670 kilometers per hour.

5. How is the orbital speed of the object determined?

The orbital speed of the object is determined by the desired orbital period, which is typically 24 hours for a geosynchronous orbit. The orbital speed can be calculated using the formula: orbital speed = (2 * pi * altitude) / orbital period.

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