1u cant conswerve energy as such as there is a non conservative viscous/drag force acting on satelite
after solving ur first differential equation we get
v=\frac{mv_0}{m-av_0t} \ \ \ \ \texttt{where} \ v_0=\sqrt{\frac{GM}{nR}}\\ \texttt{also we have this binding relation between v and r as}\\ v^2=\frac{GM}{r}\\ \texttt{putting v from first equation, we get a relation b/w r and t} \\ \frac{1}{\sqrt{\frac{nR}{GM}}-\frac{at}{m}}=\sqrt{\frac{GM}{r}}\\ \texttt{now put r as R ,i.e the time when it lands on planet,and solve for t}\\ \frac{1}{\sqrt{\frac{nR}{GM}}-\frac{at}{m}}=\sqrt{\frac{GM}{R}}\\ \texttt{we get t as} \boxed{t=\frac{m}{a}\left( \sqrt{n}-1\right)\sqrt\frac{R}{GM}}
1Then in that case you have to consider the work done by gravity on the satellite.
The satellite actually follows a spiral path, hence there is a component of gravity though small along the spiral orbit.
1
varun.tinkle
·2010-10-13 09:51:08
i think this is the best method....
first
-dv/dt=kv^{2}/m
therefore v at any time is
v=mv_{0}/ktv_{0}+m(@ bindaas the vel will decrease with time... not increase)
here v_{0}=\sqrt{Gm/nR}
and from here
we derive the equation...
R=GM/nv^{2}
therefore
-dR/dt=2Gm/nv^{2} dv/dt
solving the differential equation we get the answer
1
swordfish
·2010-10-13 11:49:24
I have a big doubt here.
The total mechanical energy should decrease due to frictional resistance.
Since the satellite is falling towards the planet, its potential energy is decreasing.
from the equations we see that v2 = √GM/R > v1 = √GM/nR
i.e. the velocity increases which means that the kinetic energy is increasing. Then how the heck will the total mechanical energy decrease???
Experts please reply