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Geostationary Orbit Energy

  • Thread starter Johfb
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  • #1
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Homework Statement



Calculate the energy of a 100kg satellite associated with a geostationary orbit around a planet.

Homework Equations



Kepler's 3rd Law:

[tex]T^{2} = \left( \frac{4 \pi^{2}}{GM_{p}} \right)r^{3}[/tex]

Velocity equation:

[tex]v = \sqrt{\frac{GM_{p}}{r}}[/tex]

Planet mass:

[tex]M_{p} = 5 \times 10^{20} kg[/tex]

Satellite mass:

[tex]m_{s} = 100 kg [/tex]

The Attempt at a Solution



Given the mass of the planet (M_{p}) I rearranged Kepler's 3rd law to get the radius of the geostationary orbit, then used the velocity equation to calculate the velocity of the satellite in this geostationary orbit.

I thought that I should then calculate the energy of the satellite by using the equation:

[tex]E = \frac{1}{2}mv^{2} - \frac{GM_{p}m_{s}}{r}[/tex]

However this gives the result of [itex]E < 0[/itex] ! :uhh:

I'm sure it's a rather simple question but I just can't figure it out at the moment.. :frown:
 

Answers and Replies

  • #2
Doc Al
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However this gives the result of [itex]E < 0[/itex] ! :uhh:
And that bothers you why? :smile:

(All bound orbits have negative energy!)
 
  • #3
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I see.. :wink:

I thought that if E<0 then it was a closed elliptical orbit and that when E=E_{min} then it was a closed circular orbit.. so I was expecting to get E_{min} which would be small but positive. I'm a tad confused.

I couldn't see where my method or calculations were incorrect, so assumed it had to be right.. but I just don't really understand why the result of the energy is negative.. perhaps could you explain? :shy:
 
  • #4
gneill
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When:

E < 0 then: bound orbit (elliptical, circular, degenerate conic (straight line trajectory))
E = 0 then: unbound orbit (parabolic or straight line; escape velocity)
E > 0 then: unbound orbit (hyperbolic or straight line; V > 0 at infinity)

Also, if e is the eccentricity of the orbit, then:

E < 0 --> 0 < e < 1
E = 0 --> e = 1
E > 0 --> e > 1
 
  • #5
gneill
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Given the mass of the planet (M_{p}) I rearranged Kepler's 3rd law to get the radius of the geostationary orbit, then used the velocity equation to calculate the velocity of the satellite in this geostationary orbit.
I'm curious to know how you determined the radius of the geostationary orbit without knowing the period of rotation of the planet.
 
  • #6
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Thanks for you explanation gneill :smile:

I'm curious to know how you determined the radius of the geostationary orbit without knowing the period of rotation of the planet.
The period of rotation of the planet is the same as the Earth i.e. T=86400s. So that's how I managed to do it :smile:
 
  • #7
Doc Al
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I thought that if E<0 then it was a closed elliptical orbit
Right. But realize that a circle is just an ellipse with zero eccentricity.

but I just don't really understand why the result of the energy is negative..
See this general derivation of the mechanical energy of an orbiting body: http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Gravity/GravityME-Dervation.html" [Broken]
 
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