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Kepler's Laws and a satellite

  • Thread starter wowdusk
  • Start date
  • #1
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Homework Statement


a satellite of mass 200 kg is launched from a site on earth's equator into an orbit 200 km above the surface of earth. a) assuming a circular orbit , what is the orbital period of this satellite? b)what is the satellite's speed in it's orbit? c) what is the minimum energy necessary to place the satellite in orbit , assuming no air friction?



Homework Equations


PE= -G*Me*m/r

how does
-γMm/(R+r) become GmR²/(R+r)


The Attempt at a Solution


I did both parts a and b just 100% stuck on part c..no idea at all what to do :(
...This makes me frustrated...really!!!
 
Last edited:

Answers and Replies

  • #2
LowlyPion
Homework Helper
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I did both parts a and b just 100% stuck on part c..no idea at all what to do :(
...This makes me frustrated...really!!!
Consider the difference in total energy of the satellite from before launch and then in orbit.
 
  • #3
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that doesnt help me much i knew that i need to use the conservation of energy...i just dont know how to apply it here................
 
  • #4
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would PEi= -GMem/ri???
like i asked before how does -γMm/(R+r)= -gmR²/(R+r)
confuzzled?!?!?!?
 
  • #5
LowlyPion
Homework Helper
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that doesnt help me much i knew that i need to use the conservation of energy...i just dont know how to apply it here................
OK. So what is the conservation of energy?

ΔPE + ΔKE = Energy supplied isn't it?
 
  • #6
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yes but would this be it .5*Ms*vi^2-GMeMs/Re=.5*Ms*vf^2-GMeMs/Rt

like what would vi be and vf be? does anything get canceled out? im not sure
 
  • #7
26
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woud vi be the speed the earth is rotating and vf be my answer from part b?
 
  • #8
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ARGGGGGGGG it still doesnt work what about -γMm/(R+r)= -gmR²/(R+r) how is this possible?!?!? idk im so confused. ive been working on this single part for over 3 hours now and still i have no answer..... i will never understand this :grumpy::mad:
 
  • #9
LowlyPion
Homework Helper
3,090
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woud vi be the speed the earth is rotating and vf be my answer from part b?
Since you don't know where it would be launched from and ... I think you can assume that the rotational speed of the Earth is sufficiently slower than orbital speed, your initial speed would be zero.

As to finding the Additional energy it's really then

E = 1/2*m*Vf2 + G*Me*Ms( 1/6380 - 1/6580)

(Pay careful attention to units. I've used km above.)
 
  • #10
26
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Thank you!!!!!! I was messing up the units XD
 

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