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

1. May 10, 2018

### 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.

2. May 10, 2018

### 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.

3. May 10, 2018

### Staff: Mentor

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.

4. May 10, 2018

### 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.

5. May 10, 2018

### Staff: Mentor

@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.

6. May 11, 2018

### olgerm

You are right.

7. May 11, 2018

### Lucw

Hello

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

8. May 11, 2018

### rcgldr

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.

9. May 11, 2018

### Staff: Mentor

olgerm assumed external forces there.

10. May 11, 2018

### olgerm

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.

11. May 11, 2018

### olgerm

if rocket has same mass as the asteroid, then $m=M$
And escaping object does push asteroid backwards, because it has rocket engine, then:
$m*v_m^2=\frac{G*m^2}{r}$
$v_m=\sqrt{\frac{G*m}{r}}$

speed is here relative to center of mass.

Last edited: May 11, 2018
12. May 11, 2018

### sophiecentaur

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.

13. May 11, 2018

### 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.

14. May 11, 2018

### Staff: Mentor

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.

15. May 11, 2018

### sophiecentaur

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.

16. May 11, 2018

### Staff: Mentor

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 think it is just adding unnecessary complication not helping OP.

17. May 11, 2018

### sophiecentaur

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?

18. May 11, 2018

### olgerm

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.

19. May 12, 2018

### 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.

20. May 12, 2018

### Staff: Mentor

Based on OP's questions, clearly the version that conserves asteroid/spacecraft momentum is easier.
Both are possible, analyzing it from the center of mass of the two objects will make the calculations easier than other choices.