# Energy required to place a satellite in orbit

1. Apr 16, 2008

### XxBollWeevilx

[SOLVED] Energy required to place a satellite in orbit

1. The problem statement, all variables and given/known data

A 200-kg satellite is placed in Earth orbit 200 km above the surface. What is the minimum energy necessary to place this satellite in orbit (assuming no air friction)?

3. The attempt at a solution

To solve this problem, I got an answer similar to the one my book says, but it is a little off and the book gives a different means of calculating the answer.

I first took the energy that the satellite would have on the Earth, which would be $$-\frac{GmM_E}{R_E}$$

I then took the energy that the satellite would have in orbit, which would be $$-\frac{GmM_E}{2(R_E+h)}$$

I then took the final minus the initial to find the difference. I received an answer of 6.43 x 10^9 J. However, the answer in my book says that I should get an answer of 6.45 x 10^9 J as given by $$\frac{mGM_E(R_E+2h)}{2R_E(R_E+h)}$$

I have no idea how to derive this formula that the book gives, and I can't see where the 2h comes from. I know that my answer is pretty close, but I would really like to know how to derive this formula. Thanks!

2. Apr 16, 2008

### Tedjn

What you have forgotten is the slight increase in potential energy of the satellite when it is 200km above the surface. Add that in and you should come up with the right answer.

3. Apr 16, 2008

### XxBollWeevilx

Doesn't $$-\frac{GmM_E}{2(R_E+h)}$$ cover the entire energy of the satellite when it is in orbit?

4. Apr 16, 2008

### alphysicist

Hi XxBollWeevilx,

If you subtract your two energy expressions I believe you do get the expression that you quoted from the book, so I think you just made an algebraic error. Perhaps you're not multiplying the h in the denominator of the orbital energy by the factor of 2?

5. Apr 16, 2008

### Tedjn

Maybe I have made a mistake, but I think that it just covers the kinetic energy. There should also be an increase in potential energy from the higher altitude, small though it is.

6. Apr 16, 2008

### alphysicist

Hi Tedjn,

I believe his expression is correct. The full energy in orbit would be, if r is the radius of the earth,

$$-\frac{G m M}{(r+h)} +\frac{1}{2}mv^2$$

and for a satellite at altitude h above the earth we can find v^2 from the centripetal force relationship:

$$v^2 =\frac{GM}{(r+h)}$$

Plugging this into the orginal gives the orbital energy $$-\frac{GmM}{2(r+h)}$$

7. Apr 16, 2008

### Tedjn

I only think that is true if you move the satellite in from infinity.

8. Apr 16, 2008

### XxBollWeevilx

I'm trying to subtract the two energy expressions but I'm either doing it completely wrong or I wasn't right in the first place.

9. Apr 16, 2008

### Tedjn

Let's see if this works out:

$$E = \Delta U \, + \, \Delta K = \left(\frac{-GMm}{r+h} \, - \, \frac{-GMm}{r}\right) \, + \, \left(\frac{1}{2}mv^2 \, - \, 0\right)$$

Because

$$\frac{GMm}{(r+h)^2} = \frac{mv^2}{r+h} \implies v^2 = \frac{GM}{r+h}$$

we have

$$\Delta K = \frac{GMm}{2(r+h)}$$

so that

$$E = \frac{GMm \cdot 2(r+h)}{2r(r+h)} \, - \, \frac{GMm \cdot 2r}{2r(r+h)} \, + \, \frac{GMm \cdot r}{2r(r+h)} = \frac{GMm(r + 2h)}{2r(r+h)}$$

Last edited: Apr 16, 2008
10. Apr 16, 2008

### XxBollWeevilx

Wow, thanks! That all pretty much makes perfect sense to me...I just get a little lost in the last line when I need to add the fractions. Could you tell me how exactly the top line would combine so as to make the 2h appear? Thank you so much for your help.

11. Apr 16, 2008

### Tedjn

Sure. Adding fractions is always annoying. I've basically made all the denominators the same and moved some terms around to get rid of double negatives and such. Then, the numerators add together using the distributive property like:

$$GMm\,\cdot\, 2(r+h) \; - \; GMm \,\cdot\, 2r \; + \; GMm \,\cdot\, r = GMm[2(r+h) \,-\, 2r \,+\, r] = GMm(r \, + \, 2h)$$

12. Apr 16, 2008

### alphysicist

Tedjn,

Were you referring to my post? If so, what part looks wrong to you?

13. Apr 16, 2008

### XxBollWeevilx

I forgot to factor out the GMm, that would have made it so much easier haha. Iget it perfectly now, thanks so much for all the help! =)

14. Apr 16, 2008

### Tedjn

alphysicist,

Most of it is correct. The only difference between mine and yours is that I started the satellite on the surface of the Earth, where there is already some potential energy. Then, moving it 200km away from (read: above) the Earth requires a little extra energy since we are raising the potential energy.

Last edited: Apr 16, 2008
15. Apr 16, 2008

### alphysicist

Tedjn,

Sorry if it seems like I'm going over details, but I'd like to get everything straightened out. My claim is that the energy of the satellite before its launched (ignoring earth's rotation) is the gravitational potential energy

$$E_i = - \frac{GmM}{r}$$

and the energy of the satellite in orbit is

$$E_f = - \frac{GmM}{2(r+h)}$$

which includes the kinetic energy plus the decreased potential energy at an altitude h. That is, I stated that I think his original expressions were exactly correct when your post stated that he did not account for the increase in potential energy. The factor of h in the equation for the orbital energy accounts for the change in the potential energy.

We can just subtract these to get the desired result:

$$E_f - E_i = - \frac{GmM}{2(r+h)} -(- \frac{GmM}{r}) = GmM\left( -\frac{1}{2(r+h)}+\frac{1}{r}\right) = GmM \left( \frac{-r + 2 (r+h) }{2 r(r+h)}\right) = GmM \frac{ (r+2h) }{2 r(r+h)}$$

It also looks to me like your approach is exactly the same. The first part of the last line in your calculation is

$$E = \frac{GMm \cdot 2(r+h)}{2r(r+h)} - \frac{GMm \cdot 2r}{2r(r+h)} + \frac{GMm \cdot r}{2r(r+h)}$$

Looking at the three terms, if you combine the second and third terms, you get

$$E = \frac{GMm 2(r+h)}{2r(r+h)} - \frac{GMm r}{2r(r+h)} = \frac{GMm }{r} - \frac{GMm }{2(r+h)}$$

which is identical to the two expressions in the original post.

Are we just having a miscommunication? Or do you see an error I made? If so, please let me know how your expression for the total energy on the ground and the total energy in orbit differ from mine.

16. Apr 16, 2008

### Tedjn

No, you're right. I just missed the negative sign. Sorry about the misunderstanding.