Is the Work-Energy Theorem Applicable to Satellite Orbital Movement?

[jwu]

Homework Statement

A satellite of mass M is in a circular orbit of radius R around the earth.

(a) what is its total mechanical energy (where Ugrav is considered zero as R approaches infinity)?

(b) How much work wouldbe required to move the satellite into a new orbit, with radius 2R?

Homework Equations

(a)
mv²/R=GMm/R² →→ mv²=GMm/R →→ K=1/2mv²=GMm/(2R),
therefore, E=K+U=GMm/(2R)+(-GMm/R)=-GMm/(2R)

(b)
Here's where I got stuck :
This is the correct answer on the book:
From the equation Ki+Ui+W=Kf+Uf,
W=(Kf+Uf)-(Ki+Ui)
=Ef-Ei
=-GMm/(2(2R))-(-GMm/(2R))
=GMm/(4R)

Here's what I did, instead of using the equation above, Ki+Ui+W=Kf+Uf, I used the WORK-ENERGY THEOREM. But it came out the different answer.

W=Kf-Ki=GMm/(4R)-GMm/(2R)=-GMm/(4R) , the same magnitude but different sign.

What's wrong with using WORK-ENERGY THEOREM?

The Attempt at a Solution

As above.

 

[tiny-tim]

Hi jwu!

Have I understood this correctly …

instead of using W = K_f - K_i + U_f - U_i,

you just used the "work-energy theorem", W = K_f - K_i ?​

ok, for the work-energy theorem, you have to include the work done by all the forces, and that includes the force of gravity, so you would have W_rocket + W_gravity = K_f - K_i … the same as the book's answer.

The only trick is that the book has replaced W_gravity by -PE.

You see, PE is just another name for (minus) work done by a conservative force (such as gravity) … you can either use work done, or you can use (minus) PE.

(btw, you have to be careful about what you regard as "energy" …

from the PF Library on potential energy …

Is potential energy energy?

There is confusion over whether "energy" includes "potential energy".

On the one hand, in the work-energy equation, potential energy is part of the work done.

On the other hand, in the conservation-of-energy equation (and conservation of course only applies to conservative forces), potential energy is part of the energy.)​

 

[bjd40@hotmail.com]

think of it:
if the distance increases how the P.E., K.E., and T.E varies?

 

[jwu]

Hi jwu!

Have I understood this correctly …

instead of using W = K_f - K_i + U_f - U_i,

you just used the "work-energy theorem", W = K_f - K_i ?​

ok, for the work-energy theorem, you have to include the work done by all the forces, and that includes the force of gravity, so you would have W_rocket + W_gravity = K_f - K_i … the same as the book's answer.

The only trick is that the book has replaced W_gravity by -PE.

You see, PE is just another name for (minus) work done by a conservative force (such as gravity) … you can either use work done, or you can use (minus) PE.

(btw, you have to be careful about what you regard as "energy" …

from the PF Library on potential energy …

Is potential energy energy?

There is confusion over whether "energy" includes "potential energy".

On the one hand, in the work-energy equation, potential energy is part of the work done.

On the other hand, in the conservation-of-energy equation (and conservation of course only applies to conservative forces), potential energy is part of the energy.)​

So basically you mean the work-energy theorem and the conservation of energy equation are interchangable at some point?

 

[tiny-tim]

So basically you mean the work-energy theorem and the conservation of energy equation are interchangable at some point?

For conserved forces (such as gravity), yes.

But for most applied forces (such as rockets, bits of string, etc), no … conservation of energy can't apply to them because, with them, energy isn't conserved.