Total energy of satellite in orbit

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SUMMARY

The discussion centers on calculating the total energy of a satellite in orbit, specifically addressing the altitude, speed, and minimum energy required to move the satellite to outer space. The satellite, with a mass of 800 kg, orbits at an altitude of 2.7 x 106 m, achieving a speed of 6.7 x 103 m/s. The minimum energy required to remove the satellite from orbit to outer space is calculated to be 1.8 x 107 J, factoring in both gravitational potential and kinetic energy. The discussion highlights a common oversight in textbook solutions regarding the inclusion of initial kinetic energy.

PREREQUISITES
  • Understanding of gravitational force equations, specifically -GMm/(r2)
  • Familiarity with gravitational potential energy calculations, -GMm/r
  • Knowledge of kinetic energy formulas, GMm/(2r)
  • Basic algebra and problem-solving skills in physics
NEXT STEPS
  • Study the derivation of gravitational force equations in orbital mechanics
  • Learn about the concept of escape velocity and its calculations
  • Explore the implications of kinetic energy in orbital transfers and maneuvers
  • Investigate the differences between radial and tangential force applications in space travel
USEFUL FOR

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

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Homework Statement


The weight of a satellite in a circular orbit round the Earth is half of its weight on the surface of the Earth. If the mass of the satellite is 800kg, determine
(i)the altitude of the satellite
(ii)the speed of the satellite in the orbit
(iii)the minimum energy required to move the satellite from its orbit to outer space.

Radius of Earth = 6.4 x 10^6 m
Mass of Earth = 6.0 x10^24 kg


Homework Equations


Gravitational force = -GMm/(r^2)
W=mg
Gravitational potential energy = -GMm/r
Kinetic energy of satellite = GMm/(2r)




The Attempt at a Solution


(i) GMm/(r^2)=mg
GMm/(r'^2)=0.5mg
Solving, r'=(2^0.5)r
Altitude,h=r'-r
=(2^0.5-1)(6.4 x 10^6)
= 2.7 x 10^6 m

(ii) GMm/(r^2)= (mv^2)/r
v=(GM/(r)^0.5
Solving using r=2^0.5 x 6.4 x 10^6 m
v = 6.7 x 10^3 m/s

(iii) Total energy of satellite in orbit is
T=U+K
When removed to outer space,energy of satellite should be U=0 and K=0 since we want to find the minimum energy required.
Thus,
Energy required,E= 0 - (U+K)
E = 0 -(-GMm/r-GMm/(2r))
= GMm/(2r)
Solving using r= (2^0.5) x 6.4 x 10^6 m and m= 800kg
E=1.8 x 10^7 J

Please check my working,especially the third part of the question. When calculating Minimum energy required to remove the satellite to outer space, we should take into account the kinetic energy of the satellite shouldn't we?
 
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I think you've got it essentially right. When calculating minimum escape energy, the idea is to include the kinetic energy due to tangential velocity and then apply force in the tangential direction. This takes a lot less energy than directing your force radially and moving out to infinity -- half as much, in fact, I think.

Your algebra looks solid.
 


fuzzywolf said:
I think you've got it essentially right. When calculating minimum escape energy, the idea is to include the kinetic energy due to tangential velocity and then apply force in the tangential direction. This takes a lot less energy than directing your force radially and moving out to infinity -- half as much, in fact, I think.

Your algebra looks solid.

Thanks. Then the answer provided by my book is incorrect. They neglected the initial kinetic energy of the satellite. Thanks for checking my work.:smile:
 

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