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Energy of a satellite

  • #1

Homework Statement



a satellite of mass 100kg travels in a circular orbit around the earth, how much energy is required to move it to an altitude of 400km?

Homework Equations



Ek=1/2*mv2
U= -G(m1m2)/r

The Attempt at a Solution


i took the satellite as the system, so as it moves to the altitude of 400km, gravity is constantly doing negative work on it, so the energy given to the satellite on the ground must be equal to the amount of energy drained by gravity's work, so it reaches the altitude.
further more, the satellite is in an orbit, so it has kinetic energy too, thus its total amount of energy needed to lift it to 400 km will be the sum of kinetic and the energy used by gravity.

for the eqn, i would write : Esum=1/2*mv2+-G(m1m2)/r

however the eqn suggests the energy of the satellite is Ek-U???
is this right? so the amount of energy required to lift the sattlite is Ek-U????
 
Last edited:

Answers and Replies

  • #2
Andrew Mason
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Is the question asking how much energy is required to move it to a circular orbit of 400 km radius from stationary launch on the surface?

AM
 
  • #4
Andrew Mason
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So it starts with:

[tex]KE_i = 0[/tex] (I will ignore earth's rotation)

[tex]U_i = -GMm/R_e[/tex]

What is its KE and U in an orbit at 400 Km about the earth's surface?

AM
 
  • #5
gneill
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I think that the question could be interpreted to mean that the satellite is initially in a circular orbit near (or at) the Earth's surface, and is to be moved to a circular orbit 400km above that.
 
  • #6
um, i think it means the energy required to move the satellite from the surface of the earth to the 400km orbit.

at 400km, the velocity of the satellite is 7689m/s so its Ek is 2.96x10^9 J
U is -5.9x10^9 J
if i use Esum=1/2*mv^2+-G(m1m2)/r

then the overall energy will be negative....
 
  • #7
Andrew Mason
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um, i think it means the energy required to move the satellite from the surface of the earth to the 400km orbit.

at 400km, the velocity of the satellite is 7689m/s so its Ek is 2.96x10^9 J
U is -5.9x10^9 J
if i use Esum=1/2*mv^2+-G(m1m2)/r

then the overall energy will be negative....
Careful. You need to find the CHANGE in U. What is the change in potential energy (as r increases, potential energy gets less negative, so the change is positive - it increases).

AM
 
  • #8
so the term -G(m1m2)/r in the equation Esum=1/2*mv^2+-G(m1m2)/r
is the CHANGE OF U????

yes it is indeed positive, change in U= 3.7x10^9

so now its Ek plus the CHANGE IN U???
 
  • #9
Andrew Mason
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so the term -G(m1m2)/r in the equation Esum=1/2*mv^2+-G(m1m2)/r
is the CHANGE OF U????

yes it is indeed positive, change in U= 3.7x10^9

so now its Ek plus the CHANGE IN U???
What is the initial U? What is the U at an altitude of 400 km? What is the difference? That is the change in U.

The change in KE is easy. It starts out with 0 and it ends up with mv^2/2. All you have to know is how to calculate v for a given orbit.

AM
 
  • #10
im confused, so how would you find the energy needed to lift the satellite?
is it the change in Ek+change in U then?
 
  • #11
gneill
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im confused, so how would you find the energy needed to lift the satellite?
is it the change in Ek+change in U then?
The change in U corresponds to the work required to lift the satellite to the required height. Keeping it there requires changing the KE so that it will orbit at that height and not just fall straight back down!
 
  • #12
Andrew Mason
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im confused, so how would you find the energy needed to lift the satellite?
is it the change in Ek+change in U then?
Well, to lift the satellite 400 km the energy is just the change in potential energy. To keep it in orbit at that altitude requires kinetic energy, which requires a different analysis.

AM
 

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