IPHO classical mechanics: A mass falls on an exoplanet

[TanWu]

Homework Statement

Suppose that a mass $M$ falls on a exoplanet with acceleration of gravity $g$. The mass is released from at a height $m$ above the surface. We define the coordinate system where the mass is realised, and use a typical rectangular coordinate system that is positive upwards and to the right. That is, $m_i = 0~\hat m$, $\hat m > 0$ upwards, and $\hat m_{\pi/2}> 0$ rightwards. When the mass $M$ falls, it acted upon my a non-conservative drag force which has the form $\vec F_n = n(m'(t))^M \hat m$ where $M$ is the mass, and $n$ is a constant. The task of this problem is to apply Newton Second Law in the $\hat m$-direction and to use energy conservation to derive the expression for Newton II. You may assume that the coordinate system we have defined for this problem is inertial, that the mass is realised from rest and that any other drag force beside the one mentioned on this planet is negligible .

Relevant Equations

$\vec F_n = n(m'(t))^M \hat m$

Attempt:

I assume that the position of the mass $M$ after it is realised its position is given by the position vectors from the origin,

$\vec m = -m(t)~\hat m$ if $m(t) > 0$

or equivalently

$\vec m = m(t)~\hat m$ if $m(t) < 0$

Either one we can use for energy conservation (I am not too sure abou this). I identity the system as non-Conservative, and therefore, work done by the $\vec F_n$ is,

$W_n = -\int n(m'(t))^{M + 1} dt$ using $m'(t) = \frac{dm}{dt}$

Apply energy conservation in the vertical ($\hat m$) direction,

$\Delta V + \Delta K = W_n$

I choose to the $\vec m = -m(t)~\hat m$ if $m(t) > 0$ expression,

$-Mgm(t) - mgm_i + \frac{1}{2}M(m'(t))_f^2 - \frac{1}{2}M(m'(t))_i^2 = -\int n(m'(t))^{M + 1} dt$

$-Mgm(t) + \frac{1}{2}M(m'(t))_f^2 = -\int n(m'(t))^{M+1} dt$

I denote $m_f'(t))^2$ as $m'(t)$ for simplicity.

$-Mgm(t) + \frac{1}{2}M(m'(t))^2 = -\int n(m'(t))^{M+1} dt$

Then taking time derivatives of each side,

$-Mgm'(t) + Mm'(t)m''(t) = -n(m'(t))^{M + 1}$

$-Mgm'(t) + Mm'(t)m''(t) = -n(m'(t))^{M + 1}$

$m'(t)[-Mg + Mm''(t) + n(m'(t))^{M}]= 0$

$m'(t) = 0$ is physically impossible or $-Mg + Mm''(t) + n(m'(t))^{M} = 0$

$-Mg + n(m'(t))^{M} = -Mm''(t)$

However, from Newton II,

$-Mg \hat m + n(m'(t))^M \hat m = Mm''(t) \hat m$

$-Mg + n(m'(t))^M = Mm''(t)$

As you can probly see, there is a contradiction between the result from energy conservation and Newton II.

I express gratitude to the person who solves my doubt.

 

[haruspex]

$\vec m = -m(t)~\hat m$ if $m(t) > 0$

or equivalently

$\vec m = m(t)~\hat m$ if $m(t) < 0$

You may be confusing yourself with your notation.
First, m is defined as the release height. Don’t use it as a variable. But since you have, I'll stick with that, using $h$ for the initial height.
The release point is defined as the origin. The vertical axis is defined as positive up. If at time t the vertical coordinate is m(t) then m(0)=0 and it finishes with m=-h.
$\vec m=m\hat m$. That is true whether y is positive (as a result of having been thrown upwards at the start, maybe) or negative. It does not switch to $\vec m=-m\hat m$ according to the sign of m.

Either one we can use for energy conservation (I am not too sure abou this). I identity the system as non-Conservative, and therefore, work done by the $\vec F_n$ is,

$W_n = -\int n(m'(t))^{M + 1} dt$ using $m'(t) = \frac{dm}{dt}$

That cannot be right. Since m starts at zero and goes negative, $m'<0$. That equation would make the work done by drag positive, when clearly it must be negative.

 

[Steve4Physics]

$\vec F_n = n(m'(t))^M \hat m$ where $M$ is the mass, and $n$ is a constant

If you raise a number to some power, the power must be dimensionless (a pure number) So raising something to the power $M$, where $M$ is mass, is wrong.

The question has other issues as well IMO. It would be interesting to know where the original question comes from.