Geostationary Satellite: Height, Equator, Uses

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A geostationary satellite must orbit at approximately 35,786 kilometers above the Earth's surface to maintain a fixed position over the equator. This equatorial orbit is necessary to ensure that the satellite's orbital speed matches the Earth's rotation, resulting in a circular path rather than an elliptical one. Geostationary satellites are primarily used for communication, weather monitoring, and surveillance, offering advantages over lower orbit satellites by providing consistent coverage of specific areas. The discussion emphasizes using Newton's laws for calculations rather than relying on Kepler's laws. Overall, geostationary satellites play a crucial role in modern technology and communication systems.
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Homework Statement


It is possible to put a satellite into an orbit so that it stays in a fixed position over a point on the Earth's equator (a geostationary or geosynchronous satellite).
a) What would the be height of such a satellite above the Earth's surface?
b) Why must such a satellite orbit above the equator?
c) What are the uses of such a satellite, as compared to lower orbit satellites?


Homework Equations





The Attempt at a Solution


a) Kepler's Third Law?
b) Fatest Velocity?
c) No Idea :D
 
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Hi ride4life! :smile:
ride4life said:
a) What would the be height of such a satellite above the Earth's surface?
b) Why must such a satellite orbit above the equator?
c) What are the uses of such a satellite, as compared to lower orbit satellites?
…
a) Kepler's Third Law?
b) Fatest Velocity?
c) No Idea :D

Forget Kepler (this is the second time, isn't it?) … there's hardly any exam questions on Kepler :wink: …

a) use Newton's second law and Newton's law of gravitation (remember, you want T = 24*3600)

b) what shape would the orbit be if it wasn't on the equator?

c) google or wiki for "geostationary" :smile:
 
a) i used kepler's third law with
r=h+6.4x106, with h being the distance from the Earth's surface to the satellite
and
T=86400s
ended getting 3.59x108m
b) orbit above the equator would be a circle, elsewhere it would be an ellipse?
c) geostationary satellites move at the same speed as the earth, allowing transmitters to maintain links for periods of time, such as radio
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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