# Can space-time curvature be applied to artificial satellites.

1. May 31, 2007

### udaykumar

I do wonder if space-time curvature can be applied to artificial satellites ....
I think yes because that could be the reason why they are revolving around earth.
Doubt:But what happens if they gain velocity more than the escape velocity.
I could be conceptually wrong but if the above theory can be applied here satellite should revolve around the sun just like planets...... but don't know if it is true... If somebody knows the answer plz do post a reply.

Last edited: May 31, 2007
2. May 31, 2007

### lalbatros

Yes space time is curved near the earth.
But the curvature is very small. Nevertheless it can be observed, specially on the synchronisation of the GPS system.
The curvature is the reason of the closed orbits of satellites, even in Newtonian (flat) mechanics the orbits can be closed.

3. May 31, 2007

### udaykumar

But what if the some satellite is launched with velocity greater than escape velocity ...that means it escapes the gravity field or curvature of earth. Then i think it should be in the range of curvature of sun (because all planets beyond it are also revolving around sun) . It should revolve around the sun.. But my doubt is , Is it true that such a satellite launched with velocity greater than escape velocity going to revolve around sun as our planets do...

4. May 31, 2007

### lalbatros

Of course a satellite can escape the earth and orbit around the sun.
And it could also escape the sun and the solar system.
The pioneer probres will leave sometimes our solar system, but unfortunately the electronic onboard will not be functionning anymore.

5. May 31, 2007

### CraigD

Don't forget that if something is orbiting the Earth, then it is also orbiting the Sun. Also you curve space-time, not a fat joke, everything curves space-time according to relativity, that is what gravity is.

CraigD, AMInstP
www.cymek.com

6. May 31, 2007

### udaykumar

"Don't forget that if something is orbiting the Earth, then it is also orbiting the Sun. "

If an object is thrown with velocity greater than escape velocity ..then it doesnot orbit the earth....

7. May 31, 2007

### CraigD

Depends on what escape velocity you are refering to:
1. Escape velocity to get into orbit about the Earth
2. Escape velocity to get out of orbit of the Earth
3. Escape velocity to get out of the orbit of the Sun
4. Escape velocity to get out of the orbit of the Galaxy
5. Escape velocity to get out of the local group
6. Escape velocity to get out of the super cluster
......
you get the idea.

CraigD, AMInstP
www.cymek.com

8. May 31, 2007

### Chris Hillman

Attempted clarification

Exactly! Or more precisely: assuming you have in mind how satellite motion is treated using our gold standard theory of gravitation, general relativity (gtr), then yes, gtr treats the motion of test particles (e.g. satellite) orbiting a massive object (e.g. the Earth) like this: the world line of the satellite is a timelike geodesic in our spacetime modeling the field outside the Earth, which is a vacuum solution of gtr (usually taken to be the Kerr vacuum with mass and angular momentum having the appropriate values).

And now the caveats (in a subject this subtle, there are always caveats):

1. "the reason why": physics since Newton has not sought "the reason why" what happens in Nature happens, but has sought only a mathematical description of what happens in Nature.

2. You hardly need gtr to model satellite motion, unless this is the context of very delicate timing scenarios such as a satellite navigation system (e.g. GPS).

I don't understand what's troubling you, but have no fear: neither Newtonian gravitation nor gtr have any trouble dealing with escape velocities!

Michel, I think you slightly garbled what you probably meant to say: in Newtonian gravitation, Earth orbiting satellites move in closed elliptical orbits, whereas in gtr, they move in quasi-Keplerian orbits which are not closed since they exhibit a tiny precession of the point of closest approach or "pericenter". This effect is small but has been confirmed in many solar system observations of planets like Mercury and Venus, various asteroids, and (far more spectacular examples here!) close orbiting binary star systems with at least one partner being a neutron star.

Uday, I think Michel's point was that general relativity is not required for typical problems involving Earth-orbiting satellites, but two notable exceptions are the GPS and the Gravity Probe B experiment. In GPS, taking account of "gravitational blueshift" as well as "relativistic Doppler shift" in the satellite signals is absolutely essential, in fact, the basic clock correction (a combination of these gtr and str effects) is build into the on-board clocks. And Gravity Probe B is a very sensitive satellite-borne experiment designed to test another type of precession effect predicted in gtr, according to which the spin axis of a gyroscope orbiting the Earth will precess very slightly (this is far smaller than the very small pericenter precession effect already mentioned).

Last edited: May 31, 2007
9. May 31, 2007

### MeJennifer

Is it really that simple in GR?
If it is, then can someone tell me the escape velocity between M1 and M2?

10. Jun 1, 2007

### CraigD

V(e)=sqrt((2GM/r))

M is the object with the greater mass and r is the distance between the centers of the objects.

CraigD, AMInstP
www.cymek.com

11. Jun 1, 2007

### MeJennifer

That's only for one mass, I am talking about two masses.

Let's have a very simple case of two non-rotating balls (but if you want can also give a solution with rotating, spherical and non-spherical balls), one with a radus r1 and a mass M1 and another one with a radius r2 and a mass M2.

What's the escape velocity in general relativity?

Last edited: Jun 1, 2007
12. Jun 1, 2007

### CraigD

Actually that is the escape velocity for two masses, but the mass of the object escaping is cancled out because you are removing it from the system.

If you were asking the escape felocity of a binary system, then M = m(1) plus m(2) and r is the distance from the center of the object escaping to the center of gravity of the binary system.

Relativity doesn't play a crucial role in escape velocity until it approches the speed of light. At that point you have to consider Lorentz transformations. At some point v(e) becomes greater than the speed of light and you have a black hole.

If I am not understanding your question can you ask it in a different way?

CraigD, AMInstP
www.cymek.com

Last edited: Jun 1, 2007
13. Jun 1, 2007

### MeJennifer

That sounds very convenient but are you sure you can just cancel things out in general relativity?

Honestly I do not understand what is the difficulty in understanding this question:

Let's have a very simple case of two non-rotating balls (but if you want you can also give a solution with rotating, spherical and non-spherical balls), one with a radus r1 and a mass M1 and another one with a radius r2 and a mass M2. What's the escape velocity in general relativity?

So no test particles or ghost observers or Newtonian limits, just plain and simple general relativity.

Last edited: Jun 1, 2007
14. Jun 1, 2007

### CraigD

V(e)=sqrt((2GM/d))

were d is much less than the speed of light. The mass of the object leaving the system has no impact on the it's escape velocity. Once the escape velocity reaches the speed of light you are at the Scwartzchild(sp) radius and there is no valid escape velocity, regardless of the mass of the object, including massless objects like photons.

If I read between the lines I think what you are asking about is what does a object moving at relativistic speeds observe when exiting the system. In this case it is a question of what frame of reference are you observing from and applying the lorentz transformations for the distance traveled and time elapsed.

CraigD, AMInstP
www.cymek.com

15. Jun 1, 2007

### MeJennifer

Sorry, I think you still do not understand my question.

Perhaps this helps, think about two balls in empty space, at what velocity do they have to separate to escape from each other in terms of their radii and masses in general relativity.

Both masses leave each other, I have no idea what you mean by system here. The whole system is the two masses. Real masses, no test particles or ghost observers. And again I am not interested in Newtonian limits, for what else would be the point to ask specifically about the case of general relativity.

Last edited: Jun 1, 2007
16. Jun 1, 2007

### pervect

Staff Emeritus
The way I understand it is this: an escape velocity exists independent of direction in the Newtonian model because the Newtonian gravitational field is the gradient of a scalar potential.

There will not in general be an escape velocity independent of direction in GR. Given a specific direction in "space", though, there may or may not be an escape velocity in that direction. This would be determined by plotting the geodesics moving in that direction at various velocities, and seeing whether or not they escape (*).

I'm implicitly assuming that one has made a space-time split in order to define the concept of a direction in space (I don't see how to get rid of this assumption).

For instance, inside the photon sphere of a black hole, there will be some directions in which one cannot escape, no matter how fast one moves (because light itself cannot escape if it starts out in that direction - inside the photon sphere even light cannot orbit the black hole for example), while there will be other directions in which one can escape if one has sufficient velocity.

(*) There's no guarantee that if an object escapes at velocity v, that it will also escape at velocity v' > v as far as I know, but I'm not aware of any specific examples where this could happen.

Last edited: Jun 1, 2007
17. Jun 1, 2007

### MeJennifer

It seems to me that unless there exists a formula or some method that demonstratively converges to a solution, and again I am not talking about test masses, ghost observers, or Newtonian limits, that the claim that general relativity has no trouble with escape velocities is a bit premature.

First of all it seems to me that the notion of an escape velocity depends on the global spacetime structure. In a closed universe and zero lambda an escape velocity cannot not even exist. In an open universe it can exist, but then what is the formula or method that demonstratively converges to a solution?

Do I seriously misunderstand the matter?

Last edited: Jun 1, 2007
18. Jun 1, 2007

### CraigD

Yes, I think you are trying to pull to much out of the problem, and it just isn't there. The escape velocity equation remains the same, except at high velocities you need to perform lorentz transformations on the equation. You have to look at the escape velocity from one of the particles frame of reference, so it's mass doesn't count towards the systems(being the object you are trying to escape from) mass so that is why you ignore it.

Escape velocity simply means that you are moving away from the other object fast enough that it's gravity can no longer hold you in an orbit. You never escape the pull of the other objects gravity, gravity has an infinate reach, but is very weak, and approaches zero at infinity.

Closed or open universe your gravitational pull will always effect every other object in the universe.

CraigD, AMInstP
www.cymek.com

19. Jun 2, 2007

### MeJennifer

You cannot ignore it, any movement between two (or more) masses causes the gravitational field to change, the spacetime is not stationary.

Yes, but in the first case there exists no escape velocity since the masses cannot stay removed from each other in the infinite future, while in the second case it does exist, the masses can stay removed from each other in the infinite future.

Last edited: Jun 2, 2007
20. Jun 2, 2007

### CraigD

Your misunderstanding the term escape velocity. The escape velocity is not the velocity required to escape the gravitational influence of another body. Escape velocity is the velocity required to escape the orbit of another body. Often it is also used on Earth to indicate the velocity required to enter an orbit around the Earth. Think of escape velocity an escape from the gravity of the other body being the dominent gravitational force on your ship.

So in both a closed and open universe you are always effected by the gravity of every other object in the universe.

CraigD, AMInstP
www.cymek.com