Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 '13 at 20:58
add comment

1 Answer

up vote 2 down vote accepted

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,

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

share|improve this answer
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 $F_2=ma$. –  grjj3 Apr 30 '13 at 22:02
it's taut, but the length isn't changing, that's what I meant. –  Jim May 1 '13 at 12:40
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.