Circular Motion of Satellite

1. Feb 22, 2009

TofuManiac

1. The problem statement, all variables and given/known data
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. Relevant 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.)

And i'm having a problem with the second part of the question.

Last edited: Feb 22, 2009
2. Feb 22, 2009

gabbagabbahey

Careful, a geostationary satellite orbits at the same angular frequency (or period) as Earth's rotation, not the same velocity.

You mean mv2/r=GMm/r2 right?
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3. Feb 22, 2009

tiny-tim

Welcome to PF!

Hi TofuManiac! Welcome to PF!

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

4. Feb 23, 2009

TofuManiac

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: Feb 23, 2009