Rotation speed for a satellite

[Firben]

Homework Statement

A geostationary satellite is located at 0'N 0'E (degrees), 36000 km above a spherical Earth with radius R(earth) = 6370 km. To scan the fieldof view, the satellite rotates around its own axis(oriented parallel to the Earth's rotation axis). It records one (constant latitude) scanline per rotation with the scan mirror steering the scanline in latitude with every rotation. A scan of the complete disc from North to South is allowed to take 25 min. With additional 5 min being required for returning the scan mirror back to its start position, one complete scan can be provided every 30 min.

When the horizontal resolution of a single scan pixel at the subsatellite point is 7.5 x 7.5 km, what is the rotational speed required to archieve this scan cycle ?

Homework Equations

α_lat,lon = 2tan^-1(Resolution/R(geo))

v = wr

v = 2H/t

The Attempt at a Solution

v = 2H/t = (2*7.5)/(30 min) = 0.5 km/min

w = v/r = (0.5 km/min)/(36000 km) = 1.38*10^-5 rad/s

Which is wrong. What is it that i don't understand here ?

 

[haruspex]

I don't understand your attempt. What is the logic behind your first equation there?

I would start by finding how many rotations are needed to cover the North-South distance.

Edit: I suspect that either you have not understood the question or I haven't. It might help if you were to describe your interpretation.

 

[Firben]

I don't know what to do, if i have the e.q, αlat,lon = 2tan-1(Resolution/R(geo)) and

And if one rotation takes 1 hour. What should i do ?

 

[gneill]

Start by making a sketch showing the angle that the vertical scan must cover in order to "see" the entire image of the Earth from pole to pole. Draw the scan profile at the distance that you know the resolution of the scan. How many scan lines are required to cover that profile?

 

[Firben]

I did make a sketch of the problem
http://s716.photobucket.com/user/Pitoraq/media/Ny bitmappsbild_zps5rr4eog4.png.html

But I am not sure if i should use the following eq.
αlat,lon = 2tan-1(Resolution/R(geo))

Not sure what i should do next

 

[gneill]

Notice how the lines that you drew from the satellite to the poles cut through the circle of the Earth? That means there's some small portion that's lying outside of the scan. So those lines really should meet the Earth surface at a tangent, and not simply connect to the poles.

Sketch in the scan profile: That would consist of a circular arc centered at the satellite that touches the surface of the Earth at the sub-satellite position (where you also happen to know the resolution). What's the value of the angle ##\alpha##? What's the length of that arc bounded by the tangent lines?

 

[haruspex]

Notice how the lines that you drew from the satellite to the poles cut through the circle of the Earth? That means there's some small portion that's lying outside of the scan. So those lines really should meet the Earth surface at a tangent, and not simply connect to the poles.

The pixel area is quoted in sq km, not sq seconds of arc, so I assume that for the purposes of this problem the satellite altitude is considered great enough that we can treat the Earth as a flat disc.
@Firben, consider a rotation of the satellite during which it scans at some latitude theta. Sketch the region of the Earth scanned.

 

[gneill]

@haruspex, I'm assuming that the satellite distance r is sufficient that the profile arc can be well approximated by a concatenation of short straight line segments of length r⋅dα = 7.5 km. This because the resolution at the center of the arc is 7.5 x 7.5 km, and by following an arc of the same radial distance from the satellite the resolution will be uniform over the whole arc. I think this will be simpler to manage mathematically than the flat disk.

 

[haruspex]

@haruspex, I'm assuming that the satellite distance r is sufficient that the profile arc can be well approximated by a concatenation of short straight line segments of length r⋅dα = 7.5 km. This because the resolution at the center of the arc is 7.5 x 7.5 km, and by following an arc of the same radial distance from the satellite the resolution will be uniform over the whole arc. I think this will be simpler to manage mathematically than the flat disk.

Ok, so you are suggesting converting the 7.5km to an angle? Your r is the satellite altitude I take it.
That's a bit more complicated than my method (which only needed Earth diameter/7.5km), but it is certainly more accurate.

 

[gneill]

Ok, so you are suggesting converting the 7.5km to an angle? Your r is the satellite altitude I take it.
That's a bit more complicated than my method (which only needed Earth diameter/7.5km), but it is certainly more accurate.

Yes, that r is the 36000km "height" of the satellite. But all I need is the length of the arc, which is trivial once the subtending angle is known.

 

[Firben]

I sketched the following:
http://s716.photobucket.com/user/Pitoraq/media/RS 2_zpsppltkqph.png.html
Im not sure if i understand the question, what are they after ?

 

[haruspex]

I sketched the following:
http://s716.photobucket.com/user/Pitoraq/media/RS 2_zpsppltkqph.png.html
Im not sure if i understand the question, what are they after ?

The drawing is a little inaccurate. Where the tangent touches, what is the angle between the tangent and the radius? What is the distance from the satellite to the centre of the earth? What does that give you for alpha?

 

[gneill]

Im not sure if i understand the question, what are they after ?

They want to know how fast the satellite needs to rotate in order to completely map the observed surface area in 25 minutes. The mapping is done in horizontal strips, the width of which is dictated by the resolution. You should try to determine how many such strips will be required to complete the image.

 

[Firben]

I remade my sketch:
http://s716.photobucket.com/user/Pitoraq/media/Ny bitmappsbild_zpst76der4z.png.html
Is it a that they want ? and then see how long it takes to traverse its distance ?

 

[gneill]

Here's my sketch:

Find the value of ##\alpha##. Find the length of the arc ab. How many "slices" of width 7.5 km does it take to sum up to the length of the arc? That's how many rotations are required to scan the arc. You need to complete that many rotations in the specified time.

 

[Firben]

If i use the following eq.

α = 2*tan^-1(3.75/36000) ≈ 0.0119°
and
s = rα = 36000*0.0119 ≈ 428.4 km

It takes 57 slices to sum up the length of the arc (428.4/7.5) = 57.12

then v = 57.12km/30 min = 1.9 rad/s

Is this right ?

 

[gneill]

If i use the following eq.

α = 2*tan^-1(3.75/36000) ≈ 0.0119°

No, that's the angular size of one "slice", not the angular extent of the whole arc (from a to b on the diagram that I posted above).

and
s = rα = 36000*0.0119 ≈ 428.4 km

You've used an angle in degrees. That won't work in the formula arclength = radius x angle. You must use radians. The calculation you've specified would return the width of the slice in kilometers, not the length of the entire arc.

It takes 57 slices to sum up the length of the arc (428.4/7.5) = 57.12

then v = 57.12km/30 min = 1.9 rad/s

Is this right ?

Nope. Find the angle subtending the whole arc from a to b, then the arc length from that. Then find out how many "slices" fit the whole arc. Note that 30 minutes is not the scan time; it includes the 5 minutes required to re-position the apparatus to begin again.

 

[Firben]

How can i find the angular size of the entire arc a and b ? I don't know the angle α, if I am using s = rα . I don't have much information to after

 

[gneill]

How can i find the angular size of the entire arc a and b ? I don't know the angle α, if I am using s = rα . I don't have much information to after

Look at the diagram. Are there any right-angled triangles that include α for which you know some side lengths?

 

[Firben]

If I am using α = tan^-1(R_e/(R_e+R_Geo) is that what you mean ?

 

[gneill]

If I am using α = tan^-1(R_e/(R_e+R_Geo) is that what you mean ?

That's the idea, but you've chosen the wrong trig function. Note where the right angle is. Which side is the hypotenuse?

 

[Firben]

I got α to be equal to 81.45° ≈ 1.42 rad
and
s=rα = 36000*2*1.42 = 102240 km
and the number of slices in 102240 km is 13632

v = 13632/25min = 545.4 km/s

Im missing something ?

 

[gneill]

If I am using α = tan^-1(R_e/(R_e+R_Geo) is that what you mean ?

That's what I mean. The tan function choice is incorrect. Look at the diagram and identify which side is the hypotenuse.

 

[Firben]

I switched side and got α to be equal to 81°,α = tan-1((Re+Rgeo)/(Re)

 

[gneill]

No! "tan" is wrong. You don't have the opposite and adjacent sides of the triangle. Which sides do you have?

 

[Firben]

But i only know the side that is R_e and the other R_e+R_Geo

 

[gneill]

But i only know one side that is R_e

Then how did you write R_e/(R_e + R_geo) ?

Look again at the diagram. I've highlighted the triangle in question in red. Of the three sides SC, SM, and CM you have values for both SC and CM. From the point of view of angle α which sides are they (Hypotenuse, Opposite, Adjacent)? What ratio can you form and what trig function does it represent?

 

[Firben]

If I am using Sin(α) = R_e/(R_e+R_Geo) i got α to be ≈ 8.646°
Is this right ?

 

[gneill]

If I am using Sin(α) = R_e/(R_e+R_Geo) i got α to be ≈ 8.646°
Is this right ?

Yes, that looks much better. Be sure to use radians when you calculate the arc length.

 

[Firben]

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 ?

 

[gneill]

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"?

 

[Firben]

v = 1448.65/25 = 58 rad/s
Is this approachable ? or is I am missing something ?

 

[gneill]

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"?

 

[Firben]

1448.65 is km or dimensionless and 25 is minutes. I am not sure what unit i should use here

 

[gneill]

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.