Answer to Question #96807 in Mechanics | Relativity for Sofea

Question #96807
Human will be heading for Mars by 2030. There are already increasing number of
unmanned missions to date, with all methods of landing on Mars have required an aero-shell
and parachute sequence, by knowing the details of Mars atmosphere and surface. Consider
a Mars lander (as particle) of mass m whose motion starts from rest at an altitude ro in a
constant Mars gravitation field gM. If a resisting force F = -kmv2 is encountered, calculate the
distance s the lander falls in accelerating from zero velocity to v. Use this analysis to discuss
about one particular failed landing which resulted in the lander Schiaparelli crashing to the
surface at 340 mph due to miscalculation of Mars descent.
1
Expert's answer
2019-10-23T09:44:11-0400

According to Newton's second law:


"m\\frac{dv}{dt}=mg-kmv^2,\\\\\n\\space\\\\\n\\frac{dv}{dt}=g-kv^2,\\\\\n\\space\\\\\n\\frac{1}{g-kv^2}dv=dt,\\\\\n\\space\\\\\n\\int \\frac{1}{g-kv^2}dv=\\int dt,\\\\\n\\space\\\\\nt=\\frac{\\text{tanh}^{-1}\\bigg(v\\sqrt{k\/g}\\bigg)}{\\sqrt{gk}},"

"v=\\sqrt{\\frac{g}{k}}\\cdot\\text{tanh}\\big(t\\sqrt{gk}\\big)."


Just that simple. This equation describes how velocity depends with time.

But we have to find how the velocity depends on distance covered. Assume that the body is being acted upon by force of gravity "P". Use the same first equation:


"F_{net}=P-kmv^2,\\\\\n\\space\\\\\n\\frac{P}{g}v\\frac{dv}{dx}=P-kmv^2."

Now put that


"a^2=\\frac{P}{km}=\\frac{g}{k}."

Then


"v\\frac{dv}{dx}=g\\bigg(1-\\frac{v^2}{a^2}\\bigg),"

"-\\frac{vdv}{a^2-v^2}=-\\frac{g}{a^2}dx,"

integrate:


"\\text{ln}(a^2-v^2)=-2(g\/a^2)x+C."

The initial conditions: "r_0=r_0, v_0=0".


"C=\\text{ln}a^2+\\frac{2gr_0}{a^2}."

Therefore, the distance the lander falls in accelerating from zero velocity to "v" is:


"x=\\frac{a^2}{2g}\\bigg(\\text{ln}a^2+\\frac{2gr_0}{a^2}-\\text{ln}(a^2-v^2)\\bigg),\\\\\n\\space\\\\\nx=\\frac{a^2}{2g}\\bigg(\\text{ln}a^2-\\text{ln}(a^2-v^2)\\bigg)+r_0,\\\\\n\\space\\\\\nx=\\frac{1}{2k}\\bigg[\\text{ln}\\frac{g}{k}-\\text{ln}\\bigg(\\frac{g}{k}-v^2\\bigg)\\bigg]+r_0."

The calculations are quite difficult, so we can understand why that mistake at 340 mph happened.


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