Solution 2.6 - Revision history

Ian at 17:41, 12 March 2010

Ian — Fri, 12 Mar 2010 17:41:46 GMT

←Older revision		Revision as of 17:41, 12 March 2010
Line 20:		Line 20:

	<math>{{m}_{1}}=</math>		<math>{{m}_{1}}=</math>
-	<math>\text{5}\textrm{.}\text{98}\times \text{1}0^{\text{24}}\text{ }</math>	+	<math>\text{5}\textrm{.}\text{98}\times \text{1}0^{\text{24}}\text{ kg}</math>
-	kg	+

-	<math>{{m}_{2}}=300\ kg</math>	+
		+	<math>{{m}_{2}}=300\ \text{kg}</math>

	The radius of the earth is		The radius of the earth is

Ian at 14:23, 12 March 2010

Ian — Fri, 12 Mar 2010 14:23:47 GMT

←Older revision		Revision as of 14:23, 12 March 2010
Line 38:		Line 38:

	<math>\begin{align}		<math>\begin{align}
-	& F=\frac{\left( 6.67\times {{10}^{-11}} \right)\times \left( 5.98\times {{10}^{24}} \right)\times 300}{{{\left( 13.37\times {{10}^{6}} \right)}^{2}}} \\	+	& F=\frac{\left( 6\textrm{.}67\times {{10}^{-11}} \right)\times \left( 5\textrm{.}98\times {{10}^{24}} \right)\times 300}{{{\left( 13\textrm{.}37\times {{10}^{6}} \right)}^{2}}} \\
-	& =\frac{119.66\times {{10}^{15}}}{178.76\times {{10}^{12}}}=0.669\times {{10}^{3}}\ =669\ \text{N} \\	+	& =\frac{119\textrm{.}66\times {{10}^{15}}}{178\textrm{.}76\times {{10}^{12}}}=0\textrm{.}669\times {{10}^{3}}\ =669\ \text{N} \\
	\end{align}</math>		\end{align}</math>

Ian: New page: Use the theory of gravitation equation $F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}}$ where, $G=6\textrm{.}67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-...$

Ian — Thu, 11 Mar 2010 20:02:49 GMT

New page: Use the theory of gravitation equation <math>F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}}</math> where, <math>G=6\textrm{.}67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-...

New page

Use the theory of gravitation equation

<math>F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}}</math>

where,

<math>G=6\textrm{.}67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-2}}</math>

In this case

<math>{{m}_{1}}</math>
is the mass of the earth and
<math>{{m}_{2}}</math>
is the mass of the satellite.

<math>d</math>
is the distance of the satellite from the centre of the earth and thus is the sum of the radius of the earth and its height above the earth.

Thus

<math>{{m}_{1}}=</math>
<math>\text{5}\textrm{.}\text{98}\times \text{1}0^{\text{24}}\text{ }</math>
kg

<math>{{m}_{2}}=300\ kg</math>

The radius of the earth is

<math>\text{6}\textrm{.}\text{37}\times \text{1}0^{\text{6}}\text{ m }</math>

and the height of the satellite above the earth in SI units is <math>7\times {{10}^{6}}\ \text{m}</math>

which gives

<math>d=</math><math>\text{13}\textrm{.}\text{37}\times \text{1}0^{\text{6}}\text{ m }</math>

Inserting all these quantities in the gravitation equation

<math>\begin{align}
& F=\frac{\left( 6.67\times {{10}^{-11}} \right)\times \left( 5.98\times {{10}^{24}} \right)\times 300}{{{\left( 13.37\times {{10}^{6}} \right)}^{2}}} \\
& =\frac{119.66\times {{10}^{15}}}{178.76\times {{10}^{12}}}=0.669\times {{10}^{3}}\ =669\ \text{N} \\
\end{align}</math>

Solution 2.6 - Revision history

Ian at 17:41, 12 March 2010

Ian at 14:23, 12 March 2010

Ian: New page: Use the theory of gravitation equation F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}} where, G=6\textrm{.}67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-...

Ian: New page: Use the theory of gravitation equation $F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}}$ where, $G=6\textrm{.}67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-...$