The atmosphere braking a satellite
Take 20 minutes to prepare this exercise.
Then, if you lack ideas to begin, look at the given clue and start searching for the solution.
A detailed solution is then proposed to you.
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An artificial terrestrial satellite (S) moves with a velocity \(\vec v\) (in the inertial geocentric frame of reference) on low orbit (the altitude \(z\) is very small compared to the radius of the Earth) and forces of friction caused by the atmosphere are exerted on it.
We write this force as :
\(\vec F = - (\pi {a^2}\mu ){v^2}\frac{{\vec v}}{v}\)
Where \(\mu\) is the density of the atmosphere and \(a\) the radius of the spherical satellite. \(m\) is the satellite mass.
If the altitude is \(z<<R_T\) :
\(\mu (z) = \mu (0)\exp ( - z/H)\)
And \(\mu(0)\) and \(H\) are constant (\(H\approx 8,5\;km\)).
Because of this force \(\vec F\), we consider the motion of the satellite (S) to be a circle, the radius of which changes slowly.
Question
Give \(z(t)\), using :
\(\tau = mH/(2\pi {a^2}\mu (0){R_T}\sqrt {{g_0}{R_T}} )\)
Where \(g_0\) is the value of the gravitational field on the ground.
\(z_i\) is the initial altitude of the satellite.
Indice
Find the velocity and the mechanical energy of a circular motion.
When there are frictions, the trajectory is almost a circular, at least for one period.
Solution
Let us suppose that the motion of (S) is a circular orbit of radius \(r(t)\) around the earth. \(r(t)\) is a slow function of time \(t\).
Mechanical energy can be used to find a relation :
\({v^2} = \frac{{G{M_T}}}{r} = {g_0}\frac{{R_T^2}}{r}\)
And :
\({E_m} = - \frac{1}{2}\frac{{Gm{M_T}}}{r} = - \frac{1}{2}\frac{{m{g_0}R_T^2}}{r}\)
With :
\(r=R_T+z\) and \({g_0} = G{M_T}/R_T^2\) (gravitational field on the ground).
The power of the friction force is :
\(P = \vec F.\vec v = - \pi {a^2}\mu (z){v^3}\)
It is equal to the variation of mechanical energy of the satellite :
\(d{E_m}/dt = P\)
We know :
\(\frac{{d{E_m}}}{{dt}} = \frac{{d{E_m}}}{{dr}}\frac{{dr}}{{dt}} = \frac{1}{2}\frac{{m{g_0}R_T^2}}{{{r^2}}}\frac{{dz}}{{dt}}\)
It follows :
\(\frac{1}{2}\frac{{m{g_0}R_T^2}}{{{r^2}}}\frac{{dz}}{{dt}} = - \pi {a^2}\mu (z){v^3}\)
So :
\(\frac{{m{g_0}R_T^2}}{{{r^2}}}\frac{{dz}}{{dt}} = - 2\pi {a^2}\mu (z){\left( {{g_0}\frac{{R_T^2}}{r}} \right)^{3/2}}\)
It follows :
\(\mu (z) = \mu (0)\exp ( - z/H)\)
\(\frac{1}{{\sqrt r }}\exp (z/H)\,dz = - \frac{{2\pi {a^2}\mu (0)}}{m}{R_T}\sqrt {{g_0}} \,dt\)
If we note :
\(\tau = mH/(2\pi {a^2}\mu (0){R_T}\sqrt {{g_0}{R_T}} )\)
The previous equation becomes :
\(\sqrt {\frac{{{R_T}}}{r}} \exp (z/H)\,dz = - \frac{{2\pi {a^2}\mu (0)}}{m}{R_T}\sqrt {{g_0}{R_T}} \,dt = - \frac{H}{\tau }dt\)
Since \(z<<R_T\) :
\(\sqrt {\frac{{{R_T}}}{r}} = \sqrt {\frac{{{R_T}}}{{{R_T} + z}}} \approx 1\)
then :
\(\exp (z/H)\,dz\, = - \frac{H}{\tau }dt\)
If \(z_i\) is the initial altitude, the altitude \(z\) for a time \(t\) is :
\(\int_{{z_i}}^z {\exp (z'/H)\,dz} ' = - \frac{H}{\tau }t\)
It follows :
\(\exp (z/H) - \exp ({z_i}/H) = - \frac{1}{\tau }t\)
Finally :
\(\exp (z/H) = \exp ({z_i}/H) - \frac{1}{\tau }t\)
Question
Numerical application :
Give the time of the fall \(t_{fall}\) from the altitude \(z_i=180\;km\) until \(z_f=0\), knowing :
\(\mu (0) = 1,3\;kg.{m^{ - 3}}\;;\;H = 8\;500\;m\;;\;a = 2\;m\;;\;{g_0} = 9,8\;m.{s^{ - 2}}\)
\({R_T} = 6\;370\;km\;;\;M = {10^3}\;kg\)
Solution
The time of the fall is:
\({t_{fall}} = ({e^{{z_i}/H}} - 1)\,\tau \approx \tau \;{e^{{z_i}/H}}\).
With \(\tau = 5\;\mu s\), we have \(t_{fall}\approx 7\;870\;s\approx 2\;h\;11\;min\).
The velocity of the satellite is almost constant (hence, \(r\approx R_T\)) and :
\(v = \sqrt {{g_0}R_T^2/r} = \sqrt {{g_0}{R_T}} = 7,9\,km.{s^{ - 1}}\)