# Tension of rope in the gravitational field of earth

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?

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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

## 1 Answer

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,

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

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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. –  Jims Bond May 1 '13 at 12:40