How Do You Calculate the Height and Latitude of a Geostationary Satellite?

TofuManiac
Messages
3
Reaction score
0
1. Homework Statement
This is an question from (NZ) NCEA Scholarship Physics practise paper.
Calculate the height of geostationary satellite and hence calculate the lattidue of a satellite dish that must be placed vertical to point at it.
2. Homework Equations
Earth's rotational velocity = 465.1 m/s
therefore v of satellite = 465.1 m/s
using Fc = Fg
(mvsqr)/r = (GMm)/r
vsqr = GM/r
r = GM/vsqr

3. The Attempt at a Solution
Working so far from me:
geostationary satellite = satellite that orbits at of Earth's rotational velocity.
Earth's rotational velocity = 465.1 m/s
therefore v of satellite = 465.1 m/s
using Fc = Fg
(mvsqr)/r = (GMm)/r
vsqr = GM/r
r = GM/vsqr
= 1.8439x10pwrof9
therefore height of geostationary satellite from surface = r - radius of earth
= 1.8439x10pwrof9 - 6.39x10pwrof6
= 1.8375x10pwrof9 m
= 1.84 x 10pwrof9 m (3s.f.)
(I'm not very sure about the answer, it's quite rediculous)

And I'm having a problem with the second part of the question.
Please help :3
 
Last edited:
Physics news on Phys.org
TofuManiac said:
geostationary satellite = satellite that orbits at of Earth's rotational velocity.
Earth's rotational velocity = 465.1 m/s
therefore v of satellite = 465.1 m/s
Careful, a geostationary satellite orbits at the same angular frequency (or period) as Earth's rotation, not the same velocity.


using Fc = Fg
(mvsqr)/r = (GMm)/r
You mean mv2/r=GMm/r2 right?:wink:
[/QUOTE]
 
Welcome to PF!

TofuManiac said:
… hence calculate the lattidue of a satellite dish that must be placed vertical to point at it.

Hi TofuManiac! Welcome to PF! :smile:

For the second part, start by saying in words what the diagram would look like.

Remember that if the dish is vertical, then the direction of the satellite must be horizontal :wink:
 
Thanks man, I totally forgot the most important fact about the geostationary satellite XD

Working:
so period of orbit = 24hr = 86400s
using derived formula - T = sqroot((4 pi^2 r^3)/(GM))
r comes down to 42250474.3m
therefore height of satellite from ground = 42250474.3 - 6.37x10^6 = 3.60x10^7m (3s.f.)

To Tiny-tim = thanks for the help too XD
Cheers :3
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top