From Mechanics

Revision as of 17:19, 13 March 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (16:16, 3 February 2011) (edit) (undo) Ian (Talk | contribs)
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	The Law of gravitation applied to a particle on Mars gives,		The Law of gravitation applied to a particle on Mars gives,
-	<math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math>	+	<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,
-	<math>a=\frac{G{{m}_{1}}}{{{d}^{2}}}</math>	+	<math>a=\frac{G{{m}_{1}}}{{{d}^{\, 2}}}</math>
	The radius of Mars must be expressed in SI units.		The radius of Mars must be expressed in SI units.

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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}