Sun's Sphere of Influence

[Columbia42]

What is the radius of the sun's sphere of influence? I can't find it anywhere and I would appreciate some help. Thanks!

 

[MeDiCS]

Heliosphere.

 

[garyw]

Also it depends what you mean by sphere of influence.
Gravitionally it's potentially universe wide.
Light wise, it's billions of light years.
Solar wind wise - as above.
I've heard estimates that there are objects in orbit of the sun up to 1ly out.

 

[SiberianTiger]

You can only regard a point in space being within a gravitational body's sphere of influence if there's a much more massive body the first body orbits around.

Assuming that you are talking about the Sun's sphere of influence in Galactic frame of reference, the formula would be (very roughly assuming that the rest of Galaxy concentrated in the core and the Sun's orbit circular):

rsoi = a sun * (m sun / m Gal) ^ (2/5) = 2.5×10^17 km * ( 1 / 7.0×10^11 ) ^ (2/5) = 4.57×10^12 km,

or 30,547 AU,

or 773.7 Pluto orbit's semi-major axes away

 

[Urwumpe]

The hill sphere of the sun is about one light-year large.

 

[SiberianTiger]

The hill sphere of the sun is about one light-year large.

By the way, my wild guessed figure is 0.48 of a light-year.

 

[Columbia42]

Thanks everyone. Let me be a little more specific. I have a plan to launch a small satelite to another star. By my calculations if the satelite travels at a constant speed of ~300 km/s it will reach Alpha Centauri in 1000 years. The problem is that once you reach solar escape velocity (42.1 km/s) and shut down the engines you start losing velocity due to the pull of the sun's gravity and by the time you reach the heliopause your velocity is down to something like 10 or 20 km/s and it will take tens of thousands of years to reach another star. Therefore my plan is to execute another burn beyond the sun's gravitational SOI bringing the spacecraft's speed up to 300 km/s. Since it's now past the gravitational pull of the sun the spacecraft will maintain that speed throughout the trip. So at what distance should I execute this burn?

 

[Urwumpe]

So at what distance should I execute this burn?

Forget it quickly - you use your fuel more economic, if you include this hyperbolic excess velocity in your escape maneuver from the sun.

Just remember the formula for hyperbolic periapsis velocity:

[math]v_{ej} = \sqrt{2 v_{c}^2 + v_{hx}^2}[/math]

v_ej is eject_burn dv
v_c is circular orbit velocity [math]v_c = \sqrt{\frac{GM}{r}}[/math]
v_hx is excess velocity.

 

[Tex]

As you leave the Suns gravitational influence, I would think the velocity of the vehicle would then be affected by the next gravity source you are close to, which should then increase your speed. I'm just making my best guess here though TBH.

 

[garyw]

Once you leave the major gravitational influence of the Sun I'd guess that the next major gravitational influence is that of the supermassive black hole at the center of the Galaxy... until the gravitional influence of another star is greater than that of the black hole.

A question I have - how much delta-v has Voyager lost during its cruise?

 

[Urwumpe]

A question I have - how much delta-v has Voyager lost during its cruise?

Well, first of all, what you wanted to hear: From the trajectory it is hard to tell, since it also gained velocity by swing-bys, but the velocity loss should be precisely the parabolic velocity at Earth:

[math]\sqrt{2} * v_E = \sqrt{2} \cdot 29.6 \frac{km}{s} = 41.8 \frac{km}{s}[/math]

(Remember, the definition of parabolic orbit is an orbit, that has absolutely no excess velocity on leaving the gravity well)

Second, what you don't want to hear: It is velocity, not velocity change. The velocity change could never be lost, since it is always in each maneuver of the probe, an investment into the future.

 

[C3PO]

As you leave the Suns gravitational influence, I would think the velocity of the vehicle would then be affected by the next gravity source you are close to, which should then increase your speed. I'm just making my best guess here though TBH.

The Sun doesn't have a SoI in Orbiter (like all the other bodies). The SoI would depend on the gravity of the black hole in the center of our galaxy, and that's not modeled in Orbiter ....... yet

(At least that's how I understand the question)

 

[Columbia42]

Forget it quickly - you use your fuel more economic, if you include this hyperbolic excess velocity in your escape maneuver from the sun.

Just remember the formula for hyperbolic periapsis velocity:

[math]v_{ej} = \sqrt{2 v_{c}^2 + v_{hx}^2}[/math]

v_ej is eject_burn dv
v_c is circular orbit velocity [math]v_c = \sqrt{\frac{GM}{r}}[/math]
v_hx is excess velocity.

I must be doing something wrong here. So the circular orbit velocity is Earth's orbital velocity (29.87) because that is the spacecraft's orbit also, right? That squared and multiplied by 2 is 1773.6968. Excess velocity is about 257 km/s because the speed I want is 300 minus the initial escape velocity of 42.1 km/s. That squared is 66512.41. The square root of the two added together is about 261. That can't be right because the velocity I want is 300 so the solution should be something considerably more than that. Should I include the escape velocity in my excess velocity?

EDIT: Never mind I figured it out. Eject burn dv should be about 303 km/s. Thanks Urwumpe!

 

[Urwumpe]

I must be doing something wrong here. So the circular orbit velocity is Earth's orbital velocity (29.87) because that is the spacecraft's orbit also, right? That squared and multiplied by 2 is 1773.6968. Excess velocity is about 257 km/s because the speed I want is 300 minus the initial escape velocity of 42.1 km/s. That squared is 66512.41. The square root of the two added together is about 261. That can't be right because the velocity I want is 300 so the solution should be something considerably more than that. Should I include the escape velocity in my excess velocity?

No, your calculation is wrong - 300² = 90,000, that is the velocity you want to have outside Sol:

[math] v_{ej} = \sqrt{2 \cdot 29.87^2 + 300.0^2} \frac {km}{s}= \sqrt{1784.43 + 90000} \frac{km}{s}= \sqrt{91784.43} \frac{km}{s}= 302.96 \frac{km}{s}[/math]

Notice the difference?

That eject velocity is not dV, but the velocity you want to reach at the end of the eject maneuver relative to the sun.

If you would do the burn afterwards, the dV would be:

[math] \Delta{v} = (\sqrt{2}-1) \cdot 29.87 \frac {km}{s}+ 300.0 \frac{km}{s} = 312.27 \frac{km}{s}[/math]

 

[n72.75]

The edge of the SOI is where the gravitational strength of the minor body(sun) is equal to the gravitational strength of the major body(galaxy).

 

[Columbia42]

No, your calculation is wrong - 300² = 90,000, that is the velocity you want to have outside Sol:

[math] v_{ej} = \sqrt{2 \cdot 29.87^2 + 300.0^2} \frac {km}{s}= \sqrt{1784.43 + 90000} \frac{km}{s}= \sqrt{91784.43} \frac{km}{s}= 302.96 \frac{km}{s}[/math]

Notice the difference?

That eject velocity is not dV, but the velocity you want to reach at the end of the eject maneuver relative to the sun.

If you would do the burn afterwards, the dV would be:

[math] \Delta{v} = (\sqrt{2}-1) \cdot 29.87 \frac {km}{s}+ 300.0 \frac{km}{s} = 312.27 \frac{km}{s}[/math]

Yes, I noticed my mistake. I meant eject velocity, not dv.