Escape velocity when the rocket's mass is not small compared to the asteroid

[Lucw]

Hello

When one calculates the speed of liberation of an object with respect to a star, one takes into account the force exerted by the star on the object. But not the force exerted by the object on the star. For small objects, this is valid. But what is the speed of release of a 1000 kg rocket that wants to leave an asteroid also 1000 kg?

Lucw.

 

[gleem]

By speed of release I assume you mean escape velocity. For the Earth it is about 11.2 Km/sec. For a tiny 1000 Kg ( 0.5 m³ ) asteroid it is about 0.365 mm/sec. The Gravitational force is extremely weak. It takes truly massive objects (like a planet) to create a significant amount of force on another body. The force between two 1000 Kg objects whose centers of masses are .about 1 m apart is about 67 micro Newtons. To give you a perspective 1 ml of water has a gravitational force (weight) from the Earth of 10,000 micro Newtons. As far as the force of the rocket exhaust on the asteroid is concerned it would be huge compared to the gravitational force. Besides accelerating the rocket the exhaust initially would help push the asteroid away from the rocket. A typical 1000 Kg Earth launched rocket would have a thrust in excess of 10,000 Newtons at sea level. If the rocket's used another method of thrust so that the exhaust did not impinge the asteroid then because of the gravitational attraction the asteroid will ever so slightly be dragged by the rockets leaving.

 

[mfb]

You can use conservation of energy to calculate it. It is a bit higher than the escape velocity of a very small object from the same asteroid if you consider the relative velocity of the two objects, and lower if you consider the velocity relative to the center of mass.

 

[olgerm]

If you assume that escaping object gets pull form some other object and does not push asteroid backwards, then
$\frac{m*v^2}{2}=\frac{G*m*M}{r}$⇒$v=\sqrt{\frac{G*M*2}{r}}$

But if it does push asteroid backwards, then:
$\frac{M*v_M^2}{2}+\frac{m*v_m^2}{2}=\frac{G*m*M}{r}$
$M*v_M=m*v_m$, because of conservation of inertia.

 

[mfb]

@olgerm: The first equation would also require that we “anchor” the asteroid in space (prevent it from accelerating due to gravity). Otherwise some energy will stay kinetic energy as you made the center of mass move.

 

[olgerm]

@olgerm: The first equation would also require that we “anchor” the asteroid in space (prevent it from accelerating due to gravity). Otherwise some energy will stay kinetic energy as you made the center of mass move.

You are right.

 

[Lucw]

Hello

Thanks you for your answers.
I need time to think.
It is not clear for me.
But i don't see any solution to lauch an object without leaning on another.
And in this case, at the end, one object is at infinity in one side. And the other at infinity in the other side...

Lucw

 

[rcgldr]

The first equation would also require that we “anchor” the asteroid in space (prevent it from accelerating due to gravity). Otherwise some energy will stay kinetic energy as you made the center of mass move.

Assuming this is a closed two body system (no external forces), the center of mass of the system does not accelerate. If the initial center of mass is used as a frame of reference, then the center of mass of the system never moves. Assuming a rocket powered object is one of the two bodies, then the exhaust plume of the spent fuel needs to be included as part of the two body system.

 

[mfb]

Assuming this is a closed two body system (no external forces), the center of mass of the system does not accelerate.

olgerm assumed external forces there.

 

[olgerm]

Assuming this is a closed two body system (no external forces), the center of mass of the system does not accelerate. If the initial center of mass is used as a frame of reference, then the center of mass of the system never moves.

If escaping object gets pull from some other object and that other object(and anchor) is part of system, then the center of mass of system does not accelerate.
If escaping object gets pull from some other object and that other object (and anchor) is not part of system, then it is not closed system.

 

[olgerm]

what is the speed of release of a 1000 kg rocket that wants to leave an asteroid also 1000 kg?

if rocket has same mass as the asteroid, then $m=M$
And escaping object does push asteroid backwards, because it has rocket engine, then:

$\frac{M*v_M^2}{2}+\frac{m*v_m^2}{2}=\frac{G*m*M}{r}$
$M*v_M=m*v_m$, because of conservation of inertia.

$m*v_m^2=\frac{G*m^2}{r}$
$v_m=\sqrt{\frac{G*m}{r}}$

speed is here relative to center of mass.

 

[sophiecentaur]

And escaping object does push asteroid backwards, because it has rocket engine, then:

That is not a practicable scenario. It would be only for the first few metres of separation that the rocket engine could be pushing against the asteroid. You would have to involve a Gun mechanism or a spring to ensure that the accelerating force on the craft is equal and opposite to the force on the asteroid. Momentum would be conserved.

Here's another scenario to contemplate. Attach your asteroid to a distant, very powerful rocket by a long rope. Use the rocket to pull the asteroid away from the much lighter craft. The asteroid would have to be accelerated enough to leave the craft behind or the craft would just stay on the ground and follow the asteroid. The magnitude of the necessary acceleration would be the same as the acceleration needed to do the separation conventionally. The Energy required to do it this way would be much more because the KE is being given to a much larger mass. Momentum would not be conserved in this experiment as there would be an external force involved.

 

[olgerm]

Escape velocity means the initial speed that is enough to escape from asteroid if no more force is applied(except gravitational pulling it back) after reaching initial speed. escape velocity must be reached at distance r (height) from center of mass of asteroid.

 

[mfb]

It would be only for the first few metres of separation that the rocket engine could be pushing against the asteroid.

Meters of separation? Chemical rockets reach escape velocity from our 1000 kg asteroid (0.2 mm/s at r=1m) within micrometers (or, in practice, enough thrust would be produced before the part contacting the spacecraft actually leaves it) and even a weak ion thruster would reach it within millimeters.

 

[sophiecentaur]

Meters of separation? Chemical rockets reach escape velocity from our 1000 kg asteroid (0.2 mm/s at r=1m) within micrometers (or, in practice, enough thrust would be produced before the part contacting the spacecraft actually leaves it) and even a weak ion thruster would reach it within millimeters.

I don't quite understand your point. You seem to be implying that the ejecta from a rocket would have no effect on the ground, once it had 'left'. Would it not have exactly the same momentum to impart to the asteroid as it did to the craft (until it had spread out so as not to hit the surface).
I still think that my idea of providing the separating force on the craft in another way would be a better model.

 

[mfb]

You seem to be implying that the ejecta from a rocket would have no effect on the ground, once it had 'left'.

No, and I have no idea how you got that impression.

You talked about rocket ejecta not hitting the asteroid after meters of separation, but there is basically no way to get separated by meters without reaching escape velocity before already.

I still think that my idea of providing the separating force on the craft in another way would be a better model.

I think it is just adding unnecessary complication not helping OP.

 

[sophiecentaur]

You talked about rocket ejecta not hitting the asteroid after meters of separation, but there is basically no way to get separated by meters without reaching escape velocity before already.

This is a scale model experiment. The sort of engine used would clearly be commensurate with the situation. There is no specification of the power of the rocket so the time or distances involved are anyone's guess. But, apart from my "one metre" figure (which you could possibly challenge) there is still the fact that the ejecta from the rocket would (if absorbed in the asteroid surface) give the asteroid a similar magnitude of momentum as the rocket itself. Surely that needs to be included in the description unless you use a different means of propulsion. You say you are not suggesting the ejecta can be ignored but you seem to be objecting only to my one meter figure. Which is unnecessarily more complicated for the OP - the effect of the ejecta or using a rope to eliminate it?

 

[olgerm]

But, apart from my "one metre" figure (which you could possibly challenge) there is still the fact that the ejecta from the rocket would (if absorbed in the asteroid surface) give the asteroid a similar magnitude of momentum as the rocket itself. Surely that needs to be included in the description unless you use a different means of propulsion. You say you are not suggesting the ejecta can be ignored but you seem to be objecting only to my one meter figure.

My equations did assume that ejecta from rocket gives asteroid same momentum as the rocket itself(but in oppsite direction). With realistic size of asteroid, mass of asteroid and power of rocket engine it is good approximation.

 

[Lucw]

Hello

Suppose we use a spring to give the speed to the rocket. Spring that push on the asteroid ...
And as this seems to disrupt the discussion, let's give an undefined mass to the asteroid, m1. And a mass similar to the rocket, m2 (of the same order of magnitude as the asteroid ...).
For my part, I will take the problem "backwards".
Let a mass m1 which lies at infinity of a mass m2.
They are released without initial speed. Just a little push so they are no more infinitely one of the other ...
A small preliminary question.
Which xyz repository to choose for the calculations?
1 / Centered on the center of the two mass? The mass m1 is then at - infinity. And the mass m2 to + infinity.
2 / Centered on one of the two masses?

Have a nice week end.

Lucw.

 

[mfb]

Which is unnecessarily more complicated for the OP - the effect of the ejecta or using a rope to eliminate it?

Based on OP's questions, clearly the version that conserves asteroid/ spacecraft momentum is easier.

Which xyz repository to choose for the calculations?
1 / Centered on the center of the two mass? The mass m1 is then at - infinity. And the mass m2 to + infinity.
2 / Centered on one of the two masses?

Both are possible, analyzing it from the center of mass of the two objects will make the calculations easier than other choices.

 

[Lucw]

Hello Mfb.

Both are possible. OK. But i am not sure the results will be the same...
What do you think?

Lucw.

 

[mfb]

The laws of physics are the same in all reference frames. Things like the speed of an object depend on the reference frame, but questions like "will the objects escape" or (in the reverse direction) "what is the collision speed" do not - at least not if we neglect corrections from special relativity.

 

[sophiecentaur]

It seems to be usual practice when discussing space rocketry to talk in terms of Δv and that makes sense for many situations. Efficiency issues mean that you make your rocket as powerful as possible. But in this case, the two bodies can acquire similar amounts of KE and there are extra considerations - especially if you ignore the efficiency requirement. If you only accelerate the space ship (with a rope) then varying the value of the acceleration will greatly affect the behaviour of the asteroid. Accelerating the ship at less than the surface g will result in towing the asteroid behind it for ever, a brief, high acceleration will hardly disturb the asteroid. It's the equivalent of the old whipping the tablecloth away trick. Intermediate acceleration will pull the asteroid but gradually leave it behind. As a non Rocket Scientist, I find that situation the most interesting. Turning off the engine and giving a slight sideways nudge before escape has been achieved would put the two in a mutual orbit.

 

[Lucw]

Hello Sophiecentaur

Oups oups oups.
I have not still made any calculations...
It is sun in my country. And basic with tomatoes, Vegetable and salads.
But we have the third law of Newton. Action and reaction.
And it seems to me that it is impossible to accelerate one mass without an accelerating of an other one...
It seems to me...

Lucw.

 

[sophiecentaur]

Hello Sophiecentaur

Oups oups oups.
I have not still made any calculations...
It is sun in my country. And basic with tomatoes, Vegetable and salads.
But we have the third law of Newton. Action and reaction.
And it seems to me that it is impossible to accelerate one mass without an accelerating of an other one...
It seems to me...

Lucw.

N3 is alive and well in UK too. You have to apply it appropriately, though. In the case of my 'rope' suggestion there will be one force between the ship and the asteroid (g) which varies with distance. If you apply an external force on the ship (my rope), it will not affect the g force but it can separate (or not) the two, according to how large it is relative to mg. As soon as the acceleration of the ship exceeds g (relative to the COM), the ship will separate from the asteroid. There will still be a force mg between ship and asteroid but g will decrease (inverse square of distance relative to COM). The result on the asteroid's eventual motion will depend on the net impulse (∫(force on asteroid times time applied)). If the force on the rope is high enough, the time will be very short so the Impulse will be small (vanishingly) and the asteroid could be considered as remaining where is was. No violation of N3.

 

[mfb]

It seems to be usual practice when discussing space rocketry to talk in terms of Δv and that makes sense for many situations.

Usual practice in rocketry is to assume a spacecraft mass that is negligible compared to the mass of the objects orbited. Lucw is explicitly asking about a scenario where this is not the case. "Usual practice" is not applicable.

 

[sophiecentaur]

. "Usual practice" is not applicable.

Absolutely.

even a weak ion thruster would reach it within millimeters.

The sort of values of force that Google gives you for an ion thruster is in the region of 0.5N and the force between two 10kg masses is minuscule. Two 10³kg lumps of rock would be only about 0.5m in radius so the g force, when touching, would be of the order of 10^-10N. To be fair, even an ion thruster would be way over the top for such an experiment.
It may be better to think in terms of a couple of much bigger objects (rocks) to give a scenario with forces that we could get a more tangible feel for it. Let's face it, the OP is describing something nearer to an almost everyday situation involving two small spacecraft or separating two satellites from a shared launcher.
This discussion also brings into perspective the navigation precision needed during the 'landing' with the Comet 67P in the Rosetta project. That object was about 2km in size.

 

[Lucw]

Hello

If we drop two objects m1 and m2 distant from infinity, what is their energy when they collide ...
Potential energy exchanged: Ep = G.m1.m2. (1 / d-1 / infinity) = G.m1.m2 / d.
d is the distance between their centers of gravity.
And it's also their kinetic energy.
Which can be expressed with the relative speed of m1 with respect to m2. V = v1 + v2.
Ek = 1 / 2.m1.m2 / (m1 + m2). (V1 + v2) ².

And so the speed that must be given to mass 1 in relation to m2 is:
V = square root (2.G. (m1 + m2) / d).
And the energy needed: G.m1.m2 / d

What do you think?

Lucw.

 

[mfb]

Ek = 1 / 2.m1.m2 / (m1 + m2). (V1 + v2) ².

Where does that equation come from?

 

[Lucw]

Hello Mfb

With falling objects, we have the conservation of momentum (of movement quantity). m1.v1 = m2.v2.
That can be written: m1/m2=v2/v1 (1)
or m2/m1=v1/v2 (2)
We add (1) and (2). And add 2:

m1/m2+m2/m1+2 = v1/v2+v2/v1+2

Same denominator gives: (m1²+m2²+2m1.m2)/m1.m2 = (v1²+v2²+2v1.v2)/v1.v2

And v1.v2 = m1.m2. (v1+v2)²/(m1+m2)²

Ek = 1/2 (m1.v1.v1+m2.v2.v2) = 1/2 (m2.v2.v1+m1.v1.v2)= 1/2 (m1+m2)v1.v2.

And Ek = 1/2. (m1.m2/(m1+m2)) . (v1+v2)²

There you go. Agree?

Lucw

 

[mfb]

An interesting approach. Looks fine.

 

[Lucw]

Hello Mfb
In the same way:

m1.m2/(m1+m2)² = v1.v2/(v1+v2)² = a1.a2/(a1+a2)² = r1.r2/(r1+r2)².

Since

m1.r1=m2.r2 m1.a1=m2.a2 m1.v1=m2.v2

And also:

m1.m2/(m1²+m2²) = v1.v2/(v1²+v2²) = a1.a2/(a1²+a2²) = r1.r2/(r1²+r2²).

We can write the Newton's law in the form:

F= G.(m1²/r1).(m2²/r2) / (m1+m2)² = a1.m1=a2.m2
Where r1 is the distance between m1 and the center of mass of the two objects. Idem for r2

And a1= G.(m1/r1).(m2²/r2) / (m1+m2)²

But come back to the escape velocity.
In the books; it is written that the escape velocily does not depend of the mass of the object which goes away.
Is this true?
Yes. But with a little condition. Which one?

Have a nice day.

Lucw

 

[Lucw]

Well well.
If we drop from infinity a mass of 1 kg on our asteroid of 1000 kg, we can calculate its speed at impact. (1).
If we drop from infinity a mass of 100 kg on our asteroid of 1000 kg, we can calculate its speed at impact. (2).
With the formula found in the book: Square root (2.G.M/d).
(1) and (2) are the same...
With the formula :V = square root (2.G. (m1 + m2) / d).
(1) and (2) are not the same (assume that diameters of objects are the same even if mass are not the same...). And don't forget V=v1+v2.

But

If we consider now the escape velocity.
If we consider that there is a volcanic eruption on our asteroid. And that a stone of mass m is ejected, what speed must it have to reach the infinite?
Formula of the books gives: Square root (2.G.(M-m)/d). So that depends of m.
Formula V=square root (2.G. (m1 + m2) / d) become: V=square root (2.G.(m1-m+m). Where m1 is the mass of the asteroid before the stone goes away...
The energy to eject the stone depends on its mass. But the relative speed of the stone compared to the asteroid does not depend on its mass ...

Have a nice day...

Lucw.
It's time to clarify Galileo's reasoning, right?

 

[mfb]

The escape velocity is always for objects with a mass much smaller than the parent body. It is the limit of $M \pm m \approx M$, and there all the formulas become identical.

 

[Lucw]

Hello Mfb

Agree with you. But in my sphere (in a forum of a Belgium University) and in the books, we find that the escape velocity does not depend of the mass of the object that escapes...
I invite you to open an old post. About falling objects.
Duration of fall. Here on this forum. Nov 10. 2017. Janus in message 15 give a page where we can find the right formule for the duration.
The conclusions of the formula are. About the raisonning of Galileo.
Galileo is right. On Earth, ALL objects fall in the same time (Same hight). At the condition that they come from Earth.
But Aristotle is also right. Heavy objects fall faster than light objects. If they don't come from Earth.
That's all.

Nice saterday.

Lucw.

(i think they will me put away...)