Solution 2.7
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(New page: <math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math> where <math>F</math> is the force on the particle, <math>{{m}_{1}}</math> is the mass of the moon, <math>{{m}_{2}}</math> is the ...)
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Revision as of 14:11, 30 March 2009
\displaystyle F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}
where \displaystyle F is the force on the particle, \displaystyle {{m}_{1}} is the mass of the moon, \displaystyle {{m}_{2}} is the mass of the particle and \displaystyle d is the radius of the moon.
As \displaystyle F={{m}_{2}}a where \displaystyle a is the acceleration of the particle
\displaystyle a=\frac{G{{m}_{1}}}{{{d}^{2}}}
As
\displaystyle G=6.67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}
we get
\displaystyle \begin{align}
& a=\frac{6.67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}\times \text{7}.\text{38}\times \text{1}{{0}^{\text{22}}}\text{kg}}{{{\left( \text{1}.\text{73}\times \text{1}{{0}^{\text{6}}}\text{m} \right)}^{2}}} \\
& =\frac{6.67\times \text{7}.\text{38}}{\text{1}.\text{7}{{\text{3}}^{2}}\times 10}\text{m}{{\text{s}}^{-2}}=1.64\text{m}{{\text{s}}^{-2}} \\
\end{align}
