Energy for orbits

  • Thread starter ice
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  • #1
ice
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This question came up on a quiz and i got it wrong.
a)how much energy does it take to send a 1000kg object from the surface of the earth to a height of 500km above the earth's surface? b) how much energy does it take to put the same object into ORBIT at 500km above the surface of the earth.

For part A, i got -4.5*10^9 Joules, but i dont know if thats right.
part b is what i really dont understand, so any help here would be appreciated, thanks.
 

Answers and Replies

  • #2
Tide
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In part A you simply calcuated the amount of work required to lift an object to a certain height. If you simply lifted it to that height and let it go it would simply fall back to Earth. In part B, the keyword is ORBIT which means that in addition to merely lifting the object you must impart to it sufficient velocity for it to remain at that height in order for it to ORBIT the Earth. You will find that the kinetic energy requirement of orbiting is substantially greater than the requirement for simply lifting the object.
 
  • #3
ice
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but isnt Ek=1/2 Ep for all orbits? i tried using v^2 = sqrt(2GM/r), which didnt really work out.
 
  • #4
futb0l
I think for b) it is:

[tex]
ME = \frac{1}{2}mv^2 - \frac{mMG}{r}
[/tex]


[tex]
ME = \frac{mMG}{2r} - \frac{mMG}{r}
[/tex]

The Mechanical Energy = Kinetic Energy + Potential Energy
I substituted the orbital velocity for v which equals to:

[tex]
v_{orb} = \sqrt{\frac{MG}{r}}
[/tex]

You can derive this by putting the Gravitational Force to be equal to the centripetal force.

If there are any errors - feel free to comment.
 
  • #5
Tide
Science Advisor
Homework Helper
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The part that you're leaving out is that the payload starts off on the surface of the Earth with whatever potential energy it has there and (in the simplified case) NO kinetic energy. What you need to calculate is the CHANGE in both the payload's potential energy and kinetic energy.
 

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