Ian at 16:12, 3 February 2011

2011-02-03T16:12:14Z

←Older revision		Revision as of 16:12, 3 February 2011
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	Consider a particle on the surface of the planet.		Consider a particle on the surface of the planet.

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

	Thus		Thus

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

	As		As
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	<math>\begin{align}		<math>\begin{align}
-	& {{d}^{2}}=\frac{\left( 6\textrm{.}67\times {{10}^{-11}} \right)\times \left( 5\times {{10}^{20}} \right)}{{{3\textrm{.}2}^{{}}}}=\frac{33\textrm{.}35\times {{10}^{9}}}{3\textrm{.}2}=10\textrm{.}4\times {{10}^{9}}=1\textrm{.}04\times {{10}^{10}} \\	+	& {{d}^{\, 2}}=\frac{\left( 6\textrm{.}67\times {{10}^{-11}} \right)\times \left( 5\times {{10}^{20}} \right)}{{{3\textrm{.}2}^{{}}}}=\frac{33\textrm{.}35\times {{10}^{9}}}{3\textrm{.}2}=10\textrm{.}4\times {{10}^{9}}=1\textrm{.}04\times {{10}^{10}} \\
	& d=\sqrt{1\textrm{.}04}\times {{10}^{5}}=1\textrm{.}02\times {{10}^{5}}\ \text{m=102}\ \text{km} \\		& d=\sqrt{1\textrm{.}04}\times {{10}^{5}}=1\textrm{.}02\times {{10}^{5}}\ \text{m=102}\ \text{km} \\
	\end{align}</math>		\end{align}</math>

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

2010-03-12T15:02:16Z

New page: Consider a particle on the surface of the planet. <math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math> where <math>F</math> is the gravitational force on a particle on the surface of t...

New page

Consider a particle on the surface of the planet.

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

where
<math>F</math>
is the gravitational force on a particle on the surface of the planet, in other words, its weight.

<math>{{m}_{1}}</math>
is the mass of the planet,
<math>{{m}_{2}}</math>
is the mass of the particle and
<math>d</math>
is the radius of the planet.

As
<math>F={{m}_{2}}a</math>
where
<math>a</math>
is the acceleration of the particle very close to the surface of the planet,

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

Thus

<math>{{d}^{2}}=\frac{G{{m}_{1}}}{a}</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}
& {{d}^{2}}=\frac{\left( 6\textrm{.}67\times {{10}^{-11}} \right)\times \left( 5\times {{10}^{20}} \right)}{{{3\textrm{.}2}^{{}}}}=\frac{33\textrm{.}35\times {{10}^{9}}}{3\textrm{.}2}=10\textrm{.}4\times {{10}^{9}}=1\textrm{.}04\times {{10}^{10}} \\
& d=\sqrt{1\textrm{.}04}\times {{10}^{5}}=1\textrm{.}02\times {{10}^{5}}\ \text{m=102}\ \text{km} \\
\end{align}</math>

Solution 2.8 - Revision history

Ian at 16:12, 3 February 2011

Ian: New page: Consider a particle on the surface of the planet. F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}} where F is the gravitational force on a particle on the surface of t...

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