Total Energy of a Geosynchronous Satellite

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SUMMARY

The total energy of a geosynchronous satellite with a mass of 1500 kg, orbiting at a height of 325 km and an orbital speed of 5000 m/s, is calculated using the equations for kinetic energy (Ek) and potential energy (Ep). The relevant formulas are Ek = (0.5)(m)(v)^2 and Ep = (m)(g)(h). The discussion emphasizes that while the orbital speed is crucial for maintaining the satellite's position, it does not directly factor into the total energy calculation as traditionally defined in this context.

PREREQUISITES
  • Understanding of kinetic energy and potential energy formulas
  • Familiarity with gravitational acceleration variations
  • Knowledge of geosynchronous satellite mechanics
  • Basic algebra for manipulating equations
NEXT STEPS
  • Research the concept of orbital mechanics and geosynchronous orbits
  • Learn about gravitational potential energy variations at different altitudes
  • Study the derivation and application of the total mechanical energy formula
  • Explore the implications of orbital speed on satellite stability and energy calculations
USEFUL FOR

Students studying physics, aerospace engineers, and anyone interested in satellite dynamics and energy calculations in orbital mechanics.

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Homework Statement


Calculate the total energy of a geosynchronous satellite (one that orbits over a fixed spot) with a mass of 1500kg, orbiting Earth at a height of 325km with an orbital speed of 5000m/s

Homework Equations


ET=Ek+Ep
Ek=(0.5)(m)(v)^2
Ep=(m)(g)(h)

The Attempt at a Solution



Looking at the orbital velocity was confusing me, so I researched geosynchronous satellites. The orbital velocity is being used to keep it stationary, assuming the Earth is rotating at the same velocity. So it is not used in these calculations.

Now, in this lesson we have not learned the formulas of Ek= G(mp)(mo) /rp and
Ep= -G(mp)(mo)

Therefore, I bet the what the teacher wants is just Et=mgh . I would like to believe that it is not asking us to include the orbital energy.

But again Et is not a vectory quanitiy and orbital energy is possible.

Would the total energy in this case be

Et=(mo)(0.5)(5000m/s)^2 + (mo)(g)(h)
 
Physics news on Phys.org
mgh applies near the Earth's surface where g is essentially constant. Way out where the satellite is, g ain't yer usual 'g' anymore! :wink:
 

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