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In a Cavendish experiment that first measured gravity, a 0.730 kg lead ball was placed 0.230 m away from a 158 kg lead ball. How much gravitational force did that create?

 

Fg=G(m1m2)/r^2

 

G=6.67*10^-11

 Nov 29, 2016
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In a Cavendish experiment that first measured gravity, 

a 0.730 kg lead ball was placed 0.230 m away from a 158 kg lead ball.

How much gravitational force did that create?

Fg=G(m1m2)/r^2

G=6.67*10^-11

 

Let m1 = 0.730  kg

Let m2 = 158  kg

Let r = 0.230  m

Let G = \(6.67\cdot 10^{-11} \ N\cdot \frac{m^2}{kg^2}\)

 

\(\begin{array}{|rcll|} \hline F_g &=& G\ \frac{m_1\cdot m_2}{r^2} \\ F_g &=& 6.67\cdot 10^{-11} \ N\cdot \frac{m^2}{kg^2}\ \cdot \frac{0.730\ kg\cdot 158\ kg}{0.230^2\ m^2} \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{0.730\cdot 158}{0.230^2} \ N\cdot \frac{m^2}{kg^2}\ \cdot \frac{ kg^2 }{m^2} \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{0.730\cdot 158}{0.230^2} \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{115.34}{0.0529} \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot 2180.34026465 \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot 2.18034026465\cdot 10^3 \ N \\ F_g &=& 6.67\cdot 2.18034026465\cdot 10^{-11} \cdot 10^3 \ N \\ F_g &=& 6.67\cdot 2.18034026465\cdot 10^{-8} \ N \\ F_g &=& 14.5428695652 \cdot 10^{-8} \ N \\ F_g &=& 1.45428695652\cdot 10^1 \cdot 10^{-8} \ N \\ F_g &=& 1.45428695652\cdot 10^{-7} \ N \\ F_g &=& 1.45428695652\cdot 10^{-1}\cdot 10^{-6} \ N \\ F_g &=& 0.145428695652\cdot 10^{-6} \ N \\ \mathbf{F_g} & \mathbf{=} & \mathbf{0.145428695652\ \mu N } \\ \hline \end{array}\)

 

laugh

 Nov 29, 2016
edited by heureka  Nov 29, 2016
edited by heureka  Nov 29, 2016

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