Conservation of Energy for satellite in an elliptic orbit

AI Thread Summary
The discussion focuses on using conservation of energy to determine the speed of a satellite in an elliptical orbit at different points (A, B, and C). Participants emphasize the importance of calculating gravitational potential energy (GPE) and kinetic energy (KE) at these points, noting that total energy remains constant throughout the orbit. The challenge lies in accurately applying the formulas and understanding how to manipulate them without needing the satellite's mass, as it cancels out in the equations. There is confusion regarding the calculations, particularly when deriving the speed at point B, leading to incorrect results. The conversation highlights the need for careful algebraic manipulation and understanding of specific mechanical energy in orbital mechanics.
NathanLeduc1
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Homework Statement


A satellite is in an elliptic orbit around the Earth. Its speed at pedigree A is 8650 m/s. (a) Use conservation of energy to determine its speed at B. The radius of the Earth is 6380 km. Use conservation of energy to determine the speed at the apogee C.


Homework Equations


KE= 0.5mv^2
PE = (-GmM)/r


The Attempt at a Solution


Honestly, I'm completely stuck. All I have so far is:
Total energy = 1/2mv^2 - (GmM)/r
I'm assuming we need to find the energy at A and that will be equal to the energy at B and at C but I really have no idea of how to go about that.
 

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NathanLeduc1 said:
I'm assuming we need to find the energy at A and that will be equal to the energy at B and at C but I really have no idea of how to go about that.
Right.
Both r and v depend on the position, but the total energy is the same. Can you find r for those 3 points?

You'll need the mass of Earth and G, or some orbital mechanics, to solve the problem.
 
r for A would be equal to 8320km
r for B would need the pythagorean formula so 8230km^2 + 13900km^2 = r^2 so r for B would equal 16150 km.
r for C is 16460 km + 8230 km = 24960 km.
 
You have a formula for the gravitational potential energy (GPE) in terms of the distance from the centre of the Earth. Have you tried calculate a number for the GPE of the satellite at point C? This seems like the easiest thing to do to start because you know the distance.

The difficult thing about point B is that the distance from the centre of the Earth isn't marked on your diagram - but can you imagine a right angled triangle on the diagram that you could solve to get the distance?
 
I don't have the mass of the satellite though so how would I do that?
 
You don't need the mass, as the energy is always proportional to the mass. It will cancel in the calculations. You can calculate the energy per mass.
 
So would I set the two energy equations equal to each other...
KE = GPE
0.5mv^2 = -(GmM)/r
0.5v^2 = -(GM)/r
Sorry to ask such basic questions, I'm just very confused about this problem.
 
You're sort of on the right track - but I'd start be writing down how energy conservation applies to this situation in the most basic way that you can think of. I'm sure you've seen something like...

total energy before anything happens = total energy afterwards.

So the question becomes; when the satellite is at A, what types of energy does it have. All of these types of energy added together make up the left hand side of the equation. Now we can do the same thing for when we're at point B for the right hand side. Don't put any numbers in yet, just see if any terms cancel out and then try to do some algebra and rearrange until you have the unknown velocity equal to some expression made up out of a bunch of things that you do know. That should get you the velocity at B, I think.
 
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Okay, I just did a lot of work and ended up with the square root of a negative number.

At A:
KE = 0.5m(v1)^2
GPE = -(GmM)/(r1)
Total: 0.5m(v1)^2-(GmM)/(r1)

At B:
KE = 0.5m(v2)^2
GPE = -(GmM)/(r2)
Total: 0.5m(v2)^2 - (GmM)/(r2)

0.5m(v1)^2-(GmM)/(r1) = 0.5m(v2)^2-(GmM)/(r2)
0.5m(v1)^2-(GmM)/(r1) + (GmM)/(r2) = 0.5m(v2)^2
0.5m(v1)^2 - ((GmM)(r1+r2))/(r1r2)= 0.5m(v2)^2
((r1+r2)(0.5m(v1)^2-GmM))/m=(v2)^2
(v2)^2 = 2(r1+r2)(0.5(v1)-GM))

which when I plug in the values:
r1 = 8.23*10^5 m
r2 = 1.62*10^6 m
v1 = 8650 m/s
G = 6.67*10^-11 Nm^2/kg^2
ME = 5.98*10^24 kg

I get the square root of a negative number.
 
  • #10
The total energy on both sides of your "before = after" equation will always be negative as the satellite is in a bound state with the Earth. If the total energy at any point were positive then the satellite wouldn't be in a closed orbit around the Earth. It's like...

- [some stuff] = -[some other stuff]

so we can just forget about the minus signs.
 
  • #11
Ok. I took the absolute value of what I got and ended up with an answer of 4.4 * 10^10 m/s. However, the answer is supposed to be 5220 m/s. Obviously, I'm WAY off. Bummer... once again, I'm stuck. Is my general approach correct? Thanks so much for working through this with me, I really appreciate it.
 
  • #12
The general idea is right, so I think the problem is somewhere in equation transformations. You can check them via WolframAlpha, if you like. A simple analysis of the dimensions might be sufficient to see the error, too.
 
  • #13
If you denote the gravitational parameter for the Earth as ##\mu_e = GM_e##, then the total specific mechanical energy of the object in orbit is
$$\xi = \frac{v^2}{2} - \frac{\mu_e}{r}$$
for a given v and r. You have a given v and r for perigee, so you can determine ##\xi## for the object. This will be a constant for the orbit.

Specific mechanical energy is the energy per unit mass of the object in orbit (J/kg). So long as the mass of the object is much less than the mass of the object it orbits (here it's the Earth) then the equation holds and you don't need to know the actual mass of the object.

Finding the corresponding v's for the given r's is then a matter of a bit of algebra and substitution of the appropriate r's.
 
  • #14
guys,we could use GMm/Re^2=mg to get GM,then we could get the right answer
 
  • #15
Myskald said:
guys,we could use GMm/Re^2=mg to get GM,then we could get the right answer
Yes, that is a good way to determine the value of GM without having to Google it or look it up in a book. However, you are responding to a thread that is nine years old. Likely the original poster has moved on.
 
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