Solution 2.10
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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 a particle on the surface of Mars, <math>{{m}_{1}}</math> is the mass of Mars, <math>{{m}_{2}...)
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Revision as of 16:58, 13 March 2010
\displaystyle F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}
where \displaystyle F is the force on a particle on the surface of Mars, \displaystyle {{m}_{1}} is the mass of Mars, \displaystyle {{m}_{2}} is the mass of the particle and \displaystyle d is the radius of Mars.
As \displaystyle F={{m}_{2}}a where \displaystyle a is the acceleration of the particle on Mars,
\displaystyle a=\frac{G{{m}_{1}}}{{{d}^{2}}}
As
\displaystyle G=6\textrm{.}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\textrm{.}67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}\times \text{
6}\textrm{.}\text{42}\times \text{1}{{0}^{\text{23}}}\text{kg}}{{{\left( \text{3}\textrm{.}\text{4}\times \text{1}{{0}^{\text{6}}}\text{m} \right)}^{2}}} \\
& =\frac{6\textrm{.}67\times \text{7}\textrm{.}\text{38}}{\text{1}\textrm{.}\text{7}{{\text{3}}^{2}}\times 10}\text{m}{{\text{s}}^{-2}}=1\textrm{.}64\text{m}{{\text{s}}^{-2}} \\
\end{align}
