How Do You Calculate the Orbital Characteristics of a Satellite?

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SUMMARY

The discussion focuses on calculating the orbital characteristics of a 220 kg satellite in a circular orbit 640 km above Earth's surface. The satellite's speed is determined to be 7555.8 m/s, and its orbital period is calculated to be approximately 5829.3 seconds. The total energy of the satellite in orbit is calculated as -6,279,885,877 J. Additionally, the discussion addresses the effects of air resistance on the satellite's altitude and speed after 1500 revolutions, emphasizing the conservation of energy and angular momentum principles.

PREREQUISITES
  • Understanding of gravitational equations, specifically Newton's law of universal gravitation.
  • Familiarity with orbital mechanics, including concepts like orbital speed and period.
  • Knowledge of energy conservation principles in physics.
  • Basic understanding of angular momentum and its calculation in circular motion.
NEXT STEPS
  • Study Kepler's laws of planetary motion for a deeper understanding of orbital dynamics.
  • Learn about the conservation of energy in orbital mechanics, focusing on kinetic and potential energy calculations.
  • Research the effects of atmospheric drag on satellite orbits and methods to calculate altitude changes over time.
  • Explore angular momentum conservation in various physical systems, particularly in celestial mechanics.
USEFUL FOR

Aerospace engineers, astrophysicists, physics students, and anyone involved in satellite design and orbital mechanics will benefit from this discussion.

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A 220 kg satellite is in an approximatel circular orbit 640 km above the Earths surface.
a) Speed of satellite and the period of its orbit?
Vt^2 = GM/R
= (6.67 x 10^-11)(6 x 10^24) / (6.37 x 10^6 + 640000) (That's radius of Earth plus height of satellite)
= 7555.8 m/s
T^2 = (4pi^2 / GM) x R^3
= 4pi^2/ (6.67 x 10^-11)(6 x 10^24) x (7010000)^3 (again, radius of Earth plus heigh of satellite)
= 5829.3

b) Total energy of satellite in its orbit?
E = -1/2 (GMm/R)
= -1/2 (6.67 x 10^-11)(6 x 10^24)(220) / 7010000)
= 6279885877 J

c) What is the angular momentum of the satellite about the centre of the Earth?
L = Iw
L = I (vt/R)
= I (7555.8/7010000)


I'm stuck now. I am pretty sure my first few answers are wrong anyway. I don't have much grasp of the concepts of this fancy gravitational stuff, I don't think. The question goes on...

d) If the satellite loses 1.5 x 10^5 J per orbital revolution due to air resistance, determine the satellite's altitude and speed after its 1500th revolution.

e) What is the angular momentum of the satellite about the centre of the Earth after the 1500th revolution? Has angular momentum about the centre of the Earth been conserved? If not, explain what has caused the change.

Needless to say, I have no idea how to tackle these.
 
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a, First start with Keplers law (T/2pi)^2 = r^3 / GM
Well done for spotting that r is the height + radius of Earth!
The speed is simple form the circumference of the orbit and period.

b, Remember total energy is kinetic + potential.

d, Use conservation of energy.
 
A is ok. the second answer needs a unit.
for B you need the kinetic energy as well, and the equation you use for the potential energy has E = 0 as R goes to infinity, E <0 everywhere else. you probably want to have E=0 at the surface of the earth.
C angular momentum is m(v x R) (cross product of vectors) for circular motion it's simply
mvr
 

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