- #1

TanWu

- 17

- 5

- 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.

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.