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?

  • $\begingroup$ Consider the acceleration of each ball in terms of the forces and tension. The two balls should have the same acceleration. $\endgroup$
    – leongz
    Commented Apr 30, 2013 at 20:58

1 Answer 1


Gotcha covered:

$$F_1=G{M_E m\over(1.5R_E)^2}\mathbf {-T}$$ -T from upwards force of rope. $$F_2=G{M_Em\over(2.5R_E)^2}\mathbf {+T}$$ +T from downwards force of rope.

Then since the rope isn't stretching,

$$F_1~=~F_2$$ $$2T~=~G{M_E m\over(1.5R_E)^2}-G{M_Em\over(2.5R_E)^2}~=~{64\over225}mg$$ $$\therefore~T~=~{32\over225}mg$$

  • $\begingroup$ Thank you very much, sir. I now see how my question was stupid! But why you say that 'rope isn't stretching'? I thought $F_1=F_2$ is because $F_1=ma$ and $F_2=ma$. $\endgroup$
    – grjj3
    Commented Apr 30, 2013 at 22:02
  • $\begingroup$ it's taut, but the length isn't changing, that's what I meant. $\endgroup$
    – Jim
    Commented May 1, 2013 at 12:40

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