Question01
An
earth satellite of mass 200 kg lost energy slowly through atmospheric
resistance and fell from an orbit of radius 8.0 x 106 m to 7.8 x 106 m.
Calculate the changes in the potential, kinetic and total energies of
the satellite during this transition period.

Answers: ( -2.57E8 J, 1.28E8 J, -1.28E8 J ) where E denotes x10



Solution:

a)

\Delta PE\ = - G \frac{M m}{r_2}\ - ( - G \frac{M m}{r_1}\ ) \Delta PE\ = - 6.67x10^{-11} \frac{(6.0x10^{24}) (200)}{7.8x10^6}\ - ( - 6.67x10^{-11} \frac{(6.0x10^{24}) (200)}{8.0x10^6}\ ) \Delta PE\ = -2.57x10^8 J

b)

Assuming the orbits are in circular motion as it decays from r_1 = 8.0 x 10^6 to r_2 = 7.8 x 10^6

equation  G \frac{M m}{r^2}\ =  \frac{m v^2}{r}\ can be assumed to be valid to approximate this motion

Thus, \Delta KE\ =\frac{1}{2}\ m v_2^2 - \frac{1}{2}\ m v_1^2

\Delta KE\ =\frac{1}{2}\ mG \frac{M}{r_2}\ - \frac{1}{2}\ mG \frac{M}{r_1}\ 



\Delta KE\ =\frac{1}{2}\ (200)(6.67x10^{-11}) \frac{6.0x10^{24}}{7.8x10^6}\ - \frac{1}{2}\ (200)(6.67x10^{-11}) \frac{6.0x10^{24}}{8.0x10^6}\ 

\Delta KE\ = 1.28x10^8 J 

c)

since  TE = PE + KE

 TE = -2.57x10^8 + 1.28x10^8

 TE = - 1.28x10^8 J



check:

This answer  TE = - 1.28x10^8 J can also be verified from the formula relationship on page Newton's Mountain Projectile Orbits which is not intended to be memorized.








Using the simulation:








using R = 8.0X10^6 , v = 7072.8 m/s for circular motion orbit, you can
get for satellite m = 1kg, KE = 2.501X10^7 J, PE = -5.002X10^7 J and TE =
-2.501X10^7 J








using R = 7.8X10^6 , v = 7162.9 m/s for circular motion orbit, you can
get for satellite m = 1kg, KE = 2.565X10^7 J, PE = -5.131X10^7 J and TE =
-2.565X10^7 J

 Thus,  

?KE = 200(2.565X10^7 - 2.501X10^7) = 1.28X10^8 J



?PE = 200(-5.131X10^7 - (-5.002X10^7)) = -2.58X10^8 J



?TE = 1.28X10^8 + (-2.58X10^8) = 1.30X10^8 J



where the slight difference is due to carry over error from the significant figures of the computer model. 















reference:

http://commons.wikimedia.org/wiki/File:R_%3D_geo_Re_2012-10-08_1809.png#.7B.7Bint:filedesc.7D.7D