Total energy of a geosynchronous satelite

1. Mar 18, 2006

James_fl

Hi all.. Could anyone help me to help this question please? I tried my best to answer the question but the answer just does not make sense.

"Calculate the total energy of a geosynchronous satellite (one that orbits over a fixed spot) with a mass of 1.5 x 10^3 kg, orbiting Earth at a height of 325 km with an orbital speed of 5.0 x 10^3 m /2"

I used Et = Ek + Ep to answer it. But I got Et < 0.

In this chapter, we haven't learned that Ep = -GMm/r. So should I use Ep = mgh instead? I didn't use this because I believe the gravity force exerted on the satellite will be considerably less than 9.8 m/s^2 since "r" has increased quite significantly.

And another thing that made me unsure is the phrase "geosynchronous satellite". A geosynchronous satellite has a period of 24 hours = 8.64 * 10^4 s.

Using the equation: T = 2(phi)r / v

To find the orbital speed necessary to maintain its orbit, we need to know the radius:

r = r earth + height
r = (6.38*10^6) m + 325000 m
r = 6.7 * 10^3 m

v = 2(phi)r / T
v = 487.23 m/s

which is different from the given v: 5000 m/s

Is this just an error that is overlooked (i.e. the geosynchronous satellite phrase means nothing) or am I doing something wrong?

Thanks,

James

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2. Mar 19, 2006

andrevdh

The fact that your calculation for the orbital speed do not agree with the given data means that the data is not consistent with a real physical problem. A geocyncronous satellite orbits at a distance of $4.24\times 10^{7}m=42,400\ km$, which can easily be checked with Kepler's third law. The height at which the satellites is suppose to orbit suggest a problem which is referred to as a low earth orbit (LEO) for a satellite - an intermediate orbit which is used to transfer a satellite to a geosynchronous or other (eg. the moon or other planet) orbit. In such a case the period will be approx 1.5 hr. I notice that you did not square the velocity in the energy equation. The total energy of objects that are bound to the force center will be negative (they cannot escape its attraction). The total energy might be $-4.5\times 10^{10}J$. What theory did you cover in the relevant chapter?

Last edited: Mar 19, 2006
3. Mar 19, 2006

James_fl

Hi andrevdh, thanks for pointing that error (v^2). I did forget to square the v.

In previous chapters, we have learned about centripetal force (Fc = (mv^2)/2, newton's law of universal gravitation (Fg = GMm/r^2). In this chapter, upto this question, the book has only discussed the law of conservation of energy, where Et = Ek + Ep. Where the Ep = mgh.

So should I use that instead? Regarding my answer, is it possible to have a negative Et?

I know that ideally I should ask my teacher -- but I do not have any, since this is a distance education course. And I need to send my answers tomorrow so I could have my midterm mark sent to a university..

Having said that, thank you so much for helping! I can only rely on this forum and its experts to clarify my answers. Do you have any suggestion on what I should do?

4. Mar 19, 2006

Hootenanny

Staff Emeritus
Yes, it is possible and in fact nesscary that the energy is negative if an object is orbiting a mass. If there is an attractive force between two objects, then the potential energy of the two objects increase with increased distance. If the objects are an infinite distance apart, then there is no potential between them. So if the potential increases to zero, it must be negative.

The explanation is a bit rough and read I know. Have you done potential energies in atoms yet?

5. Mar 19, 2006

James_fl

Hootenanny: You mean Fg = (GMm/r^2)? I studied that in my chemistry course but I haven't reached that chapter in my physics course. But thank you for your explanation. It was very clear.

If Et is always negative, then I will just use the same equation for Ep. As andrevdh correctly stated, I will correct my the v^2 part.

Thanks guys!! :-)

6. Mar 19, 2006

andrevdh

The change in the potential energy of the satellite to raise it from the surface of the earth up to its orbit can be evaluated via $U=mgh$
by noting that
$$g=\frac{GM}{r^2}$$
$g$ will therefore be less at the orbiting position. Continuing with
$$\Deta U=U_{orbit}-U_{surface}$$
and taking the zero potential level in the center of the earth we come to
$$\Deta U=m(g_or_o\ -\ gr_e)$$
if you use the above relation for $g$ one gets the same formula as one would get with
$$U=-\frac{GMm}{r}$$
(except for the minus sign which is aresult of different reference levels)
I would think that you should include your calculation for the orbital speed and comment that it is inconsistent with the other given data - this might score you some extra points!

Last edited: Mar 19, 2006
7. Mar 19, 2006

James_fl

Sure I will! Thank you soooo much andrevdh. I really appreciate your help.

8. Mar 19, 2006

BerryBoy

If it is orbiting the Earth and always focused at the Earth, the it will be rotating with a period of 24 hours. Do you need to account for this?

Sam

9. Mar 19, 2006

James_fl

BerryBoy: I believe you do... The period of revolution will determine the speed and/or the radius needed to maintain its orbit, which affects its kinetic energy (1/2 * mv^2). The total energy would also change because the kinetic energy changes (Et = Ek + Ep).

10. Mar 20, 2006

BerryBoy

OK, I don't think you understood what I was trying to say.. Have you been given the moment of inertia for this satellite to calculate its rotational energy?

$$E_{rot} = \frac{1}{2}I\omega^2$$

11. Mar 20, 2006

BerryBoy

Not only is the satellite orbiting around about a radius (giving it kinetic energy), it will also be rotating so that it always points towards the Earth so that:

$$E_{total}=E_{pot.} + E_{kin.} + E_{rot.}$$

Sam

12. Mar 20, 2006

James_fl

BerryBoy: Good point. It actually makes sense but I don't think we have learned about "moment of inertia". All that was given is in the first post.

Is there any way to calculate the moment of inertia from the given data?

Last edited: Mar 20, 2006
13. Mar 21, 2006

BerryBoy

My apologies.

To calculate the moment of inertia, you would have to make an assumption on the satellite's shape (i.e. sphere, box or something else). For the sphere and the box and other simple shapes, you can look up the values on the internet; but it is tricky to work out the other shapes.

Sam

14. Mar 21, 2006

Hootenanny

Staff Emeritus
You are not given the dimensions of the satallite, therefore you have to assume it is a particle or point mass (mass but no volume). As it has no volume, and hence no dimensions, it has no moment of inertia and therefore has no rotational kinetic energy. So you sould just consider potential energy and linear kinetic energy as you have done.

-Hoot