What Rotational Speed is Needed for a Satellite's 30-Minute Scan Cycle?

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The discussion focuses on determining the rotational speed required for a geostationary satellite to complete a 30-minute scan cycle of the Earth. The satellite must cover a North-South distance while accounting for a horizontal resolution of 7.5 km per pixel. Participants emphasize the need to calculate the angle subtending the scan arc and the total number of scan lines required to achieve full coverage. Various equations are debated, with suggestions to convert distances into angles for accurate calculations. Ultimately, the goal is to find the angular velocity in appropriate units, leading to a clearer understanding of the satellite's scanning requirements.
  • #31
Firben said:
So if
s = rα = 36000*2*(π/180)*8.646 ≈ 10864.9 km
and the number of slices in 10865.9 is 1448.65
Thus
v = 1448.65/25 ≈ 58 km/s
Is this right ?
You're looking for an angular velocity. What units are appropriate? What units are associated with your values "1448.65" and "25"?
 
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  • #32
v = 1448.65/25 = 58 rad/s
Is this approachable ? or is I am missing something ?
 
  • #33
Firben said:
v = 1448.65/25 = 58 rad/s
Is this approachable ? or is I am missing something ?
You're almost there, but you didn't answer my last question: What units are associated with your values "1448.65" and "25"?
 
  • #34
1448.65 is km or dimensionless and 25 is minutes. I am not sure what unit i should use here
 
  • #35
Firben said:
1448.65 is km or dimensionless and 25 is minutes. I am not sure what unit i should use here
A value can't both have units of km and be dimensionless; it's a logical contradiction.

In fact, the "1448" value represents the number of slices, and one slice requires one rotation of the satellite. So an appropriate unit for it is "rotations". So your ratio is rotations/minutes, or rotations per minute, or "rpm" to use the common nomenclature.

rpm is a perfectly good unit for angular velocity, but you could also convert that to radians per second if they are looking for the answer to be expressed that way.
 
  • #36
Ok, thanks for helping me out
 

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