Take the tour ×  Two balls of mass $m$ each one are connected
with mass-less rope with the same length as the radius of earth.
The system is in free fall. Prove that the tension of the rope
when the nearest (to the earth) ball's distance from the earth
surface is $R_E/2$ is: $T = \frac{32}{225} mg$   What I did is
the following:  $F_1$ is a gravitation force exerted on the
nearest ball by the earth: $F_1=G \frac{M_Em}{(1.5R_E)^2}$
$F_2$ is a gravitation force exerted on the farthest ball by the
earth: $F_2=G \frac{M_E m}{(2.5R_E)^2}$  $T=F_1-F_2=G \frac{M_E
m}{(1.5R_E)^2}-G \frac{M_E m}{(2.5R_E)^2}=\frac{G M_E m}{R_E^2}
\left (\frac{4}{9} - \frac{4}{25} \right)=\frac{64}{225} mg$
However, my answer is somehow twice bigger than what is
expected. Where am I wrong? What am I missing?  share|improve
this question Consider the acceleration of each ball in terms of
the forces and tension. The two balls should have the same
acceleration. –  leongz Apr 30 at 20:58 add comment  1 Answer
up vote 2 down vote accepted