Surface Area of Satellite Signal transmission on Earth!

  • Thread starter ron_jay
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Hi guys. Lets say we put a satellite on a geosynchronous orbit over the earth. The signal transmitted by the satellite does not reach the exact half of the earth as much as you can intuitively try but is a portion formed by the two tangents from the point the satellite is on. The problem is to calculate the effective surface area the signal from the satellite can reach on the earth.Give it a try!Would you?
 

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  • #2
malawi_glenn
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Is this a homework question? What have you tried so far so people can help u?
 
  • #3
mgb_phys
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The easiest method is probably to ignore the Earth being spherical, just draw a triangle.
You need to know the height for geostationary orbit and the radius of the Earth to solve for the effective radius at the tangent.
As you point out it would only be 1/2 the Earth's surface if the satellite was at an infinite distance.
In practice the signal is only usable if it is more than a certain angle above the horizon. This reduces the amount of atmopshere the signal must travel through and the amount of 'ground clutter' likely to be in the way of the receiver.
 
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No! this is certainly not an assignment. This problem was born out of pure rumination. Yes, I have thought quite a bit about this problem. The idea of the critical angle and the the 'ground clutter' is novel but leaving these practicalities aside, the aim is to obtain a perfectly geometrical solution and then we can delve into the harder areas of it. To draw a triangle is correct, but we have to find the surface area and for that we need to apply a bit of calculus.Check the attachment to this post. The surface area to be found is the effective sectional segment of the sphere as shown in the diagram. I found out the effective radius(r) and the distance of the centre of the circle from the surface (d) equated to a function of R(radius of earth) and h(height of the satellite).
Now, the obstacle tingling me is that how do I calculate the surface area of that portion? Integrate it. But how? Well, its not as simple as just drawing a 'triangle'!
 

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  • #5
George Jones
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Let C be an arbitray point on the Earth's surface between N and A, and let [itex]\theta[/itex] be the angle between ON (or OS) and OC. Suppose [itex]\theta[/itex] is changed by an amount [itex]d \theta[/itex]. This results in a point C' on the Earth's surface that is a distance (as measured on the Earth's surface) [itex]Rd\theta[/itex] away from C.

Revolve C and C' around OS, resulting in a circular band of the Earth's surface that has thickness [itex]Rd\theta[/itex] and radius [itex]Rsin\theta[/itex].

Can you see this?

Thus the area [itex]dA[/itex] of this band is length (circumference of the circle) times width:

[tex]dA = \left( 2 \pi R sin\theta \lright) \left(Rd\theta \left);[/tex]

[tex]A = 2 \pi R^2 \int_0^{\theta'} sin \theta d\theta[/tex]

Here, [itex]\theta'[/itex] is the angle between ON and OA.
 
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That may have found the the area of the band till [tex]\theta[/tex], but how do we calculate the spherical area of that section?
 
  • #7
Dick
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That may have found the the area of the band till [tex]\theta[/tex], but how do we calculate the spherical area of that section?

What's the difference between 'area' and 'spherical area'???????
 
  • #8
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what i mean is that the area of the 'band' is the area of the strip and is different from the solid surface area of the section which we want to find out! One of my other questions is that can a satellite be positioned in a geosynchronous orbit anywhere apart from the equatorial radius? if yes why,if no why?What will be the net effect on the satellite quantitatively?
 
  • #9
Dick
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The area George computed was the true surface area. It may look like he's approximating when describing the bands, but it's calculus. In the limit as dr->0 the result is exact. If the satellite orbit is off the equatorial plane, then as viewed from earth the satellite will not appear stationary but will follow a loop in the sky once a day.
 
  • #10
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right that's correct about the view from earth but how do we calculate whether the satellite will stay at that partial radius orbit or not?
 
  • #11
Dick
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"partial radius orbit"? What does that mean? If you question means can an orbit be geosynchronous in a circular orbit at a different radius, then the answer is no.
 
  • #12
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Yes, that is what I mean by that.So, if it cannot be at any other radius then what will be the net effect(of forces) on the satellite if it happens to be on that orbit?Which direction shall it tend,certainly having lost its stability(HOW?)?How can we calculate the height at which height the satellite can rotate in a geosynchronous orbit(the angular speed same as that of the earth)?Is there a height range for this kind of orbit?
 
  • #13
Dick
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'In orbit' means the only force acting on it is gravity. If it's at a different radius than the geosynchronous radius it doesn't 'lose it's stability' or anything. It simply doesn't have a 24hr period. To actually compute this use Kepler's third law - the period squared of an orbiting body is proportional to the radius cubed. Sure there is a height range. You can't put anything EXACTLY someplace and even if you could it wouldn't stay there. The range depends on how much drift in position you can put up with day to day. Communication satellites have engines so they can adjust their position to compensate for this.
 
  • #14
Let C be an arbitray point on the Earth's surface between N and A, and let [itex]\theta[/itex] be the angle between ON (or OS) and OC. Suppose [itex]\theta[/itex] is changed by an amount [itex]d \theta[/itex]. This results in a point C' on the Earth's surface that is a distance (as measured on the Earth's surface) [itex]Rd\theta[/itex] away from C.

Revolve C and C' around OS, resulting in a circular band of the Earth's surface that has thickness [itex]Rd\theta[/itex] and radius [itex]Rsin\theta[/itex].

Can you see this?

Thus the area [itex]dA[/itex] of this band is length (circumference of the circle) times width:

[tex]dA = \left( 2 \pi R sin\theta \lright) \left(Rd\theta \left);[/tex]

[tex]A = 2 \pi R^2 \int_0^{\theta'} sin \theta d\theta[/tex]

Here, [itex]\theta'[/itex] is the angle between ON and OA.


can u please send a soft copy of calculation via ms word ..please
 

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