Solution 2.10 - Revision history

Ian at 16:16, 3 February 2011

2011-02-03T16:16:44Z

←Older revision		Revision as of 16:16, 3 February 2011
Line 1:		Line 1:
	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
Line 19:		Line 19:
	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.

Ian at 17:19, 13 March 2010

2010-03-13T17:19:03Z

←Older revision		Revision as of 17:19, 13 March 2010
Line 1:		Line 1:
		+	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>

Line 16:		Line 18:
	<math>a</math>		<math>a</math>
	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.

-	As	+	<math>d=3\textrm{.}4\times {{10}^{6}}\ \text{m}</math>
		+
		+	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
-

	<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}}} \\
-	& =\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}} \\	+	& =\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>

Ian: New page: $F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$ where $F$ is the force on a particle on the surface of Mars, ${{m}_{1}}$ is the mass of Mars, ${{m}_{2}...$

2010-03-13T16:58:42Z

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}...

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}}</math>
is the mass of the particle and
<math>d</math>
is the radius of Mars.

As
<math>F={{m}_{2}}a</math>
where
<math>a</math>
is the acceleration of the particle on Mars,

<math>a=\frac{G{{m}_{1}}}{{{d}^{2}}}</math>

As
<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

<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{
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}</math>

Solution 2.10 - Revision history

Ian at 16:16, 3 February 2011

Ian at 17:19, 13 March 2010

Ian: New page: F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}} where F is the force on a particle on the surface of Mars, {{m}_{1}} is the mass of Mars, {{m}_{2}...

Ian: New page: $F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$ where $F$ is the force on a particle on the surface of Mars, ${{m}_{1}}$ is the mass of Mars, ${{m}_{2}...$