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}...)
Current revision (16:16, 3 February 2011) (edit) (undo)
 
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<math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math>
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The Law of gravitation applied to a particle on Mars gives,
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<math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{\, 2}}}</math>
where
where
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is the acceleration of the particle on Mars,
is the acceleration of the particle on Mars,
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<math>a=\frac{G{{m}_{1}}}{{{d}^{\, 2}}}</math>
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<math>a=\frac{G{{m}_{1}}}{{{d}^{2}}}</math>
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The radius of Mars must be expressed in SI units.
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<math>d=3\textrm{.}4\times {{10}^{6}}\ \text{m}</math>
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As
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and as
<math>G=6\textrm{.}67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}</math>
<math>G=6\textrm{.}67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}</math>
we get
we get
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<math>\begin{align}
<math>\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{
& 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}}} \\
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}}} \\
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& =\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}} \\
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& =\frac{6\textrm{.}67\times 6\textrm{.}42}{{{3\textrm{.}4}^{2}}}\ \text{m}{{\text{s}}^{-2}}=3\textrm{.}7\ \text{m}{{\text{s}}^{-2}} \\
\end{align}</math>
\end{align}</math>

Current revision

The Law of gravitation applied to a particle on Mars gives,

\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}}}

The radius of Mars must be expressed in SI units.

\displaystyle d=3\textrm{.}4\times {{10}^{6}}\ \text{m}

and 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 6\textrm{.}42}{{{3\textrm{.}4}^{2}}}\ \text{m}{{\text{s}}^{-2}}=3\textrm{.}7\ \text{m}{{\text{s}}^{-2}} \\ \end{align}

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