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

[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...)

 

[mfb]

we find that the escape velocity does not depend of the mass of the object that escapes...

Well, it does in general:

V = square root (2.G. (m1 + m2) / d).

It depends on both masses.
We can also use $v_1 = \frac{V m_2}{m_1+m_2}$ to find $$v_1 = m_2 \sqrt{\frac{2G}{d(m_1+m_2)}}$$ which depends on both masses as well.

If we assume $m_1 \ll m_2$ then $m_1+m_2 \approx m_2$ and we get $v_1\approx V \approx \sqrt{\frac{2Gm_2}{d}}$, now the velocity does not depend on m₁ any more (as long as it is small).

 

[Dale]

Heavy objects fall faster than light objects.

No. We have been over this exhaustively before.

 

[olgerm]

No. We have been over this exhaustively before.

$\frac{d^2R}{dt^2}=-\frac{\mid F_{Earth}\mid}{M}-\frac{\mid F_{object}\mid}{m}$
$\mid F_{Earth}\mid=\frac{G*M*m}{R^2}$
$\mid F_{object}\mid=\frac{G*M*m}{R^2}$
$\frac{d^2R}{dt^2}=-\frac{G*(M+m)}{R^2}$

• R is distance between falling object and Earth.
• m is mass of falling object.
• M is mass of Earth.
• t is time.
• G is gravitational constant.

If falling speed means relative speed between Earth and falling object, then object with bigger mass falls faster.
If falling speed means speed of falling object in inertial frame of refference , then on same distance from Earth all objects have same acceleration regardless of their mass. But having fallen equal time, massive objects have bigger speed, bigger acceleration and smaller distance from Earth.

 

[Dale]

If falling speed means relative speed between Earth and falling object, then object with bigger mass falls faster.
If falling speed means speed of falling object in inertial frame of refference , then on same distance from Earth all objects have same acceleration regardless of their mass.

As was already discussed exhaustively in his previous thread on the topic.