Ship thrust required to move out of the plane of our solar system

[Albertgauss]

TL;DR Summary

ship thrust solar system plane

Hello,

We know that most celestial objects in our solar system are in the equatorial plane of the sun. So too, does most of our spacecraft orbit in this plane as it explores our solar system.

For a spacecraft already traveling away from the sun and towards the outer solar system, how hard would it be for a spacecraft to thrust out of and perpendicular to the solar system plane, either up or down, while still heading outbound? That is, if the spacecraft already has a pure, outbound radial vector, would the energy required to accumulate a perpendicular velocity vector component (to the solar system plane) be a significant fraction of the ship’s energy or really cost no extra energy at all?

I know that gravity is a radial force from basic physics, of course, so in theory there is really no physics work done moving azimuthally or in polar angle directions at some radius from the sun, but it seems some energy would certainly be required to change the ship’s motion to fly out of the solar system plane. Okay with ballpark answers on this, just looking for a general idea.

 

[phinds]

I know that gravity is a radial force from basic physics, of course...

and yet you seem to have lost track of the fact that there will BE a radial element of such movement(*). There would not be if the sun was the only body in the solar system, or possibly even if the sun and the Earth were the only bodies, but there ARE others.

* And actually, it's NOT "radial" that you want, it's "perpendicular to the plane" and that's really the force I'm pointing out.

 

[phinds]

Or, to look at it in a slightly different way, consider the case where you just magically place an object on a geodesic between the sun and the Earth and they are the only two bodies in the solar system (for simplicity). The object will travel on the geodesic and you are asking if any force is requited to move it OFF of the geodesic. Well, of course there will since even in GR, Newtons' First Law still applies.

 

[DaveC426913]

... or really cost no extra energy at all?... it seems some energy would certainly be required to change the ship’s motion to fly out of the solar system plane.

If the goal is to change the trajectory and minimize energy expenditure, then a planetary slingshot would accomplish that nicely. Instead of going around a planet in the plane of the ecliptic though, you go under its pole. When you come out, you are headed out of the PotE.

 

[Albertgauss]

@phinds, yes, still maintain the radial outbound vector. That component of the motion would not change. And the energy to keep moving outwards against the sun's gravity as the radius changes would still be the same.

Yes, assume that there are no other bodies in the solar system. That will be fine for the simplicity I am looking for here. You can even ignore the Earth.

No energy efficiency needed, just, ship going along in the plane of the solar system at speeds typical of space probes, now boost perpendicular, above or below the plane, that is, change latitude. Is this "boost" going to require a significant amount of energy compared to the energy the spacecraft already has as far as the gravity of the sun is concerned? That is, is there any gravitational energy needed to be overcome by the ship to travel in a latitude direction? The energy for the boost would come from the ship. I think from classical physics, there is no gravitational work to change latitude but I don't know what GenRel would say about that.

@phinds, yes, I agree there will be gravity to overcome by changing to another geodesic since the distance from the sun has changed, but what if the ship wants to change its azimuthal or polar directions, any gravity to overcome for that?

Sorry if a little repetition, just thinking through this.

 

[hmmm27]

There's really not much energy involved, unless you want to get to the new radial line, quickly.

Think a car in a parking lot : technically the energy is the same whether you're parked in aisle A or aisle Z, but moving from A-Z requires energy expenditure to start and stop (and there's no regeneration mechanism).

 

[mfb]

Energy is rarely a useful concept in spaceflight - it depends too much on the reference frame. Spaceflight is all about delta_v, velocity changes done by the spacecraft .

Just leaving an idealized plane needs an infinitesimal amount of delta_v. Fire the smallest thruster by the shortest possible time and you are now in a different plane.
If you have a purely radial motion then the orbital plane is ill-defined, the motion is a single line.

A more realistic scenario for a spacecraft coming from a planet is a combination of radial velocity and tangential velocity that's roughly aligned with the plane of the planets. If you want to change the orbital plane by 90 degrees (a fully polar orbit) but keep the other orbital parameters: You need to cancel the tangential velocity in the plane of the planets and add an equal amount of velocity orthogonal to that plane. That's a delta_v of sqrt(2) times the original tangential velocity. The absolute number depends on your specific trajectory. A spacecraft in an Earth-like orbit would need ~42 km/s, a spacecraft in a Jupiter-like orbit would only need ~18 km/s.

 

[Janus]

Summary:: ship thrust solar system plane

Hello,

We know that most celestial objects in our solar system are in the equatorial plane of the sun. So too, does most of our spacecraft orbit in this plane as it explores our solar system.

For a spacecraft already traveling away from the sun and towards the outer solar system, how hard would it be for a spacecraft to thrust out of and perpendicular to the solar system plane, either up or down, while still heading outbound? That is, if the spacecraft already has a pure, outbound radial vector, would the energy required to accumulate a perpendicular velocity vector component (to the solar system plane) be a significant fraction of the ship’s energy or really cost no extra energy at all?

I know that gravity is a radial force from basic physics, of course, so in theory there is really no physics work done moving azimuthally or in polar angle directions at some radius from the sun, but it seems some energy would certainly be required to change the ship’s motion to fly out of the solar system plane. Okay with ballpark answers on this, just looking for a general idea.

The general solution is for the Delta v needed to change the inclination of a spacecraft can be found here:
https://en.wikipedia.org/wiki/Orbital_inclination_change.

And while the major bodies of Solar system do orbit close to the same plane, there is enough difference for between their orbital planes that it has to be accounted for with our probes.
For example, there is a 1.85 degree difference between Earth's and Mars' orbits. This could result in a probe missing Mars by as much as 7 million km if not allowed for.

This usually involves a mid-course correction called a "Broken Plane" maneuver. The craft waits until it reaches a point where its trajectory crosses the orbital plane of the target body (called a "node".) And then does a burn which changes its orbital inclination to match that of the target body.

 

[mfb]

This usually involves a mid-course correction called a "Broken Plane" maneuver. The craft waits until it reaches a point where its trajectory crosses the orbital plane of the target body (called a "node".) And then does a burn which changes its orbital inclination to match that of the target body.

That needs too much fuel. Spacecraft going to Mars are directly sent onto an intersecting course (the orbital plane of the spacecraft is neither aligned with Earth nor with Mars in general), because it's easier to launch in a slightly different direction from Earth. Mid-flight maneuvers are then limited to course corrections because no trajectory planning is ever perfect.

 

[Albertgauss]

Lots of good information here. I think my question is answered, but if people have more information, by all means post it. It has especially helped me that you posted the wikipedia, orbital inclination site.

 

[TonyCross]

To change the heading of a spacecraft usually requires a planetry slingshot or a burn from the main engine/engines.
It is also correct that pitch and yaw can be easily changed by very small effort from it's bow thrusters, however it's like a car driving forward, then skidding on ice the car is still heading in the same direction but the front of the car is now facing a different direction.
It is possible to change the pitch or the yaw with simply throwing a heavy object across the craft, as long as you are not at the centre of mass.
Newtons 3rd comes into play as you throw the item an equal force is applied to you and the craft, as the item moves across the craft, during the period of the items travel the craft is subject to the reactive force, until it hits the far wall then the rotary displacement (angular momentum) around it's center of mass will be canceled (assuming it doesn't bounce back). Obviously this change in Yaw or pitch would be quite small, dependent on the mass of the item thrown.

 

[mfb]

The orientation (attitude) is commonly controlled by gyroscopes, but I don't see the relevance for this thread.

 

[TonyCross]

The relevance is that craft orientation is not a change of direction, just a change in pitch / yaw. Yes gyroscopes are used for these axis movements.

 

[hmmm27]

Orientation notwithstanding, the fuel burn approaches zero to change radials. Of interest (to me) is that - since centripetal force comes into play during the move - it can be done by diverting the radial thrust (treating the craft as a point source), causing an angular jerk while maintaining radial velocity.

 

[mfb]

The relevance is that craft orientation is not a change of direction, just a change in pitch / yaw. Yes gyroscopes are used for these axis movements.

But no one asked about the orientation. The thread is exclusively about the orbit.

 

[mpresic3]

I remember reading about the Ulyssees mission on the news. Apparently, the probe uses a "slingshot" around Jupiter to go out of the plane. The answer seems to be little if any thrust would be needed, just a lot of clever planning, (and waiting).