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- Thread starter ron_jay
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malawi_glenn

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Is this a homework question? What have you tried so far so people can help u?

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mgb_phys

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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'!

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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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.

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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Dick

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What's the difference between 'area' and 'spherical area'???????

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Dick

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Dick

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Dick

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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.

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

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