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

[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.