Energy Required to Launch a Geosynchronous Satellite

  • #1
23
0
Hi all,

This question is for an assignment due Friday. My friend and I spent the better part of an hour on it and are thoroughly stumped. Any help is much appreciated!

1. Homework Statement

Calculate the energy required to put a satellite into geostationary orbit. (note that no mass is provided)

Homework Equations


Derivation:

PEgravitational + KE = Etotal
-GMm/r + 1/2mv2 = Etotal

Note that when in non-inertial reference frames,

Fg = mv2/r
GmM/r2 = mv2/r
GM/r2 = v2/r
GM/r = v2

Plug this value in place of v2

-GMm/r + 1/2m(GM/r) = Etotal
-1/2GMm/r = Etotal

So the energy of the satellite once in stable orbit is -1/2GMm/r, or one half of it's potential energy.

The Attempt at a Solution


Calculate altitude of geosynchronous orbit. At geosynchronous orbit, only force on the satellite is gravity.

Fnet = ma
Fnet = GMm/r2
ma = GMm/r2
mv2/r = GMm/r2

V = dist/ time
V = 2rpi / T

m(2rpi/T)2/r = GMm/r2
m4rpi2/T2 = GMm/r2

4rpi2/T2 = GM/r2

Isolate for r:

4r3pi2/T2 = GM

r = 3√(GMT2/4pi2)
= 42 250 474m

Subtract radius of Earth to determine height of geosyncronous orbit above surface:

42 250 474metres - 6 371 000metres = 35 879 474metres

Now it should just be a matter of plugging numbers into the equation for 1/2PEg. Since I don't know the satellites mass I'll represent it with 'm' as a placeholder:

1/2PEg = -GMm/2r

= -(6.67*10-11 * 5.98 * 1024)m / 2(35 879 474)
= -5 558 414m (in juoles)

Note again that m is a placeholder for the satellite's mass.

While mathematically I can't find an error, it seems intuitively wrong that it requires negative energy, if that even exists, to launch a satellite into geosynchronous orbit. It takes a lot of energy to get it moving sideways so fast, and even though the potential energy does decrease with the rise in altitude it just seems so wrong that the whole process involves negative energy.

Can somebody break this down for me, pointing out either a) how it's possible to have negative energy required to launch a satellite or b) where I've gone conceptually or mathematically astray?

Thanks!
 

Answers and Replies

  • #2
35,268
11,535
The way you use the radius does not make sense. With that approach, putting it in an orbit one meter above the surface (ignoring atmosphere, mountains and so on) would need an insane amount of energy, and launching it to outer space would just need a very gentle push. That is unrealistic.

r, as you use it in formulas, is always relative to the center of earth. You can use it to calculate the energy of the satellite in a geostationary orbit. You'll need a different equation to find the energy of a satellite sitting on the ground, then you can take the difference.

Please use consistent units everywhere.
 
  • #3
23
0
"With that approach, putting it in an orbit one meter above the surface (ignoring atmosphere, mountains and so on) would need an insane amount of energy, and launching it to outer space would just need a very gentle push. That is unrealistic."

Yep, that's a good way to think about it. I had forgotten of that tactic - 'exaggerating to extremes' - to check the validity of an approach. So I re-tried it your way, as follows:

Step 1- calculate energy of satellite if it were 'orbiting' right at surface level.
r = 6 371 000metres

PE = -GMm/2r
PE = -6.67 * 10-11 * 5.98 * 1024m / (2 * 6 371 000)
PE = -31 303 249m

Step 2- calculate energy of satellite at full geosynchronous orbit, measuring the radius from the Earth's centre.

PE = -GMm/2r
PE = -6.67 * 10-11 * 5.98 * 1024m / (2 * 42 250 474)
PE = -4 720 254m

∆PE = P1 - P2

= 26 582 994m, where the answer is in joules and is multiplied by the mass of the satellite in kg. I'll round it off to a more reasonable number, i.e 2 or 3 significant digits. = 26 600 000m

Does that look okay?
 
Last edited:
  • #4
35,268
11,535
The satellite does not start orbiting at the surface. It starts at rest (neglecting the rotation of earth).
And I think there is a wrong number in the third line for the geostationary orbit.

And don't forget the units.
 
  • #5
23
0
D'oh good point! In orbit, it's energy is comprised of KE and PEg. PEg has been calculated, so to that must be added the KE.

KE = mv2/2
I don't have v but can re-write it as V = 2rpi / T

KE = m(2rpi/T)2/2
KE = 4mr2pi2/2T2
KE = 4m(42 250 474)2pi2/2(86400)2
KE = 4 720 254m, answer in joules

So Etotal = PE + KE
Etotal = 26 582 994m + 4 720 254m
Etotal = 31 303 248m

Interestingly, that's the same *edit* magnitude */edit* as the potential energy required to lift the satellite from the Earth's centre to geosynchronous orbit altitude, as calculated in my previous post. Perhaps PE from Earth's centre is a convenient shorthand to calculate the energy required to put up a geostationary satellite?
 
  • #6
23
0
"And I think there is a wrong number in the third line for the geostationary orbit."

Yes, you're right. I edited it to reflect the proper value.

"And don't forget the units."

Etotal = 31 303 248m J
Etotal = 31.3m MJ
 
  • #7
35,268
11,535
m has units of mass, so your result is "mass*energy", that is not an energy. You get the right units naturally if you do not remove them everywhere.

The numbers were better before and now they got messed up somehow.
The 4.7 million [what?] are indeed the kinetic energy of the satellite, but it does not make sense to add them to the number of 26.6 million [what?].
 
  • #8
23
0
"m has units of mass, so your result is "mass*energy", that is not an energy. You get the right units naturally if you do not remove them everywhere."

Perhaps my notation was confusing (as you noted, lacking units makes it difficult to work with). What I meant was the 31 300 000 value is not in energy because it's lacking a 'mass.' It's only m2/s2, so it must be multiplied by a value in kg in order to elicit energy as a unit. Hence, 31.3m is a value in Juoles.

But that is a weird notation, so I'll abandon it. The issue is that I don't have mass, so I just use 'm' as a placeholder for the mass of the satellite in kg. I should have expressed the answer as 31 300 000m2/s2 * x kg, yielding an answer in J, which can for convenience's sake re-written as MJ.

"The numbers were better before and now they got messed up somehow.
The 4.7 million [what?] are indeed the kinetic energy of the satellite, but it does not make sense to add them to the number of 26.6 million [what?]."

Hm... 4.7 million m2/s2 * x kg = KEsatellite
PEg satellite = PE1 - PE2

Where PE1 is the gravitational potential energy of the satellite with respect to the Earth's surface, and PE2 is the gravitational potential energy of the satellite with respect to the Earth's centre. Using my calculations from post #3,

PE1 = -4 720 254m2/s2 * x kg
PE2 = -31 303 249m2/s2 * x kg

∆PE = 26 582 994m2/s2 * x kg

From my class notes, the total energy of a satellite is the sum of its potential energy and its kinetic energy. Therefore:

Etotal = 31 303 2482/s2 * x kg

Another way of thinking of this value is that it takes 31.3 MJ per kilogram to launch a satellite. So just multiply the satellite's mass by 31.3MJ/kg to get total energy required. My class notes also say that the energy of an orbiting satellite is equal to -PEg/2 (derivation in initial post). Double checking with this method yields:

E total = GMm/2r
E total = (6.67 * 10-11m3/(kg * s2) * 5.98 * 1024 kg * m) / 2(42 250 474metres)
E total = 4 720 254m2/s2 * x kg

The double checking method finds that I require only 4 720 254J/kg to launch the satellite, which is at odds with my previous answer. I have to admit though that the logic in my previous answer seems more sound: find the increase in PE, find the increase in KE, and add them together to find the total answer. Could you give me some pointers to how I can improve my steps?
 
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  • #9
35,268
11,535
Perhaps my notation was confusing
and wrong.
It's only m2/s2
Correct (and equivalent to J/kg), and not J as you wrote.

The issue is that I don't have mass, so I just use 'm' as a placeholder for the mass of the satellite in kg.
If you specify "m in kg", that is possible as well. I see you did that in post 3. Okay.


In post 3 you did not calculate the potential energy. That was the total energy in an orbit, otherwise the factor 2 would be wrong there.
 
  • #10
23
0
"and wrong."

Truth hurts! Better to learn here than on a test. I understand the issue now- basically, as you said earlier, use consistent units and it will work out.

"In post 3 you did not calculate the potential energy. That was the total energy in an orbit, otherwise the factor 2 would be wrong there."

You're right! Wow, didn't see that. So the calculation:

E tot = -PEg/2
E tot = -(-GMm)/2r
E tot = GMm/2r
E tot = 6.67 * 10-11m3/(kg * s2)* 5.98 * 1024 kg (x kg) / (2 * 42 250 474 metres)
E tot = 4 720 254J/kg

yields the correct answer for the total amount of energy that the satellite has. But that's not actually the value I'm interested in; I want to know how much energy must be put into a satellite to get it into space. So what I really need to do is find a) PEg on the surface of the earth and PEg at orbiting height, b) find ∆PE and c) KE. Adding b) and c) together should give me the amount of energy required to put a satellite in geosynchronous orbit.

KE = mv2/2
KE = m(2rpi/T)2/2
KE = 4mr2pi2/2T2
KE = 4(x kg)(42 250 474 metres)2pi2/2(86400s)2
KE = 4 720 254J/kg

PEorbit = -GMm/r
PEorbit = -6.67 * 10-11m3/(kg * s2)* 5.98 * 1024 kg (x kg) / 42 250 474metres
PEorbit = - 9 440 509J/kg

PEground = -GMm/r
PEground = -6.67 * 10-11m3/(kg * s2)* 5.98 * 1024 kg (x kg) / 6 371 000 metres
PEground = -62 606 498J/kg

∆PE = PEorbit - PEground
∆PE = - 9 440 509J/kg - (-62 606 498J/kg)
∆PE = 53 165 989J/kg

E tot = ∆PE + KE
E tot = 53 165 989J/kg + 4 720 254J/kg
E tot = 57 886 243J/kg

Does that look okay?
 
  • #11
Nathanael
Homework Helper
1,650
239
Looks good to me. (My answer agrees to the first two significant figures.)
 
  • #12
gneill
Mentor
20,913
2,862
Just FYI, it'll take a little less energy if you take advantage of Earth rotation (once per day). Tthe satellite on the ground starts with some KE due to this rotation and if you choose the appropriate launch direction on the equator you can maximize its contribution.
 
  • #13
35,268
11,535
Just FYI, it'll take a little less energy if you take advantage of Earth rotation (once per day). Tthe satellite on the ground starts with some KE due to this rotation and if you choose the appropriate launch direction on the equator you can maximize its contribution.
Indeed. In terms of energy, that is a small effect, but it becomes much more important if we look at actual rockets and orbital mechanics.
 

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