Conservation of Energy for satellite in an elliptic orbit

In summary: You are trying to solve for v in terms of r and g. In order to do that, you need to know the mass and gravitational potential energy of the satellite. You also need the position of the satellite. However, you do not have the mass of the satellite. You also do not know the position of the satellite.In summary, the satellite is in an elliptic orbit around the Earth. Its speed at pedigree A is 8650 m/s. Its speed at pedigree B is 16150 km, and its speed at pedigree C is 24960 km.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
I don't have the mass of the satellite though so how would I do that?
 
  • #6
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.
 
  • #7
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.
 
  • #8
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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  • #9
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.
 

FAQ: Conservation of Energy for satellite in an elliptic orbit

What is the conservation of energy for a satellite in an elliptic orbit?

The conservation of energy for a satellite in an elliptic orbit refers to the principle that the total energy of the satellite remains constant throughout its orbit. This means that the potential energy and kinetic energy of the satellite will vary as it moves through different points in its elliptical path, but the total energy will always remain the same.

How does the conservation of energy apply to satellites in elliptic orbits?

The conservation of energy applies to satellites in elliptic orbits because the gravitational force between the satellite and the central body (such as a planet or star) is the only force acting on the satellite. This means that the total energy of the system, which includes the potential and kinetic energy of the satellite, remains constant due to the conservation of energy principle.

How does the shape of an elliptic orbit affect the conservation of energy for a satellite?

The shape of an elliptic orbit does not affect the conservation of energy for a satellite. The conservation of energy principle applies to all types of orbits, including circular and elliptic orbits. However, the specific values for potential and kinetic energy will vary depending on the shape of the orbit.

How is the conservation of energy related to the speed of a satellite in an elliptic orbit?

The conservation of energy is related to the speed of a satellite in an elliptic orbit because the kinetic energy of the satellite is directly proportional to its speed. As the satellite moves through different points in its orbit, its speed will change, but the total energy will remain constant.

What is the significance of the conservation of energy for satellites in elliptic orbits?

The conservation of energy is significant for satellites in elliptic orbits because it allows us to predict and understand the behavior of these satellites. By knowing that the total energy of the system remains constant, we can calculate the potential and kinetic energy at different points in the orbit and make predictions about the satellite's motion and velocity. This principle also helps us to design and maintain efficient orbits for satellites.

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