Solution 2.8

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(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...)
Current revision (16:12, 3 February 2011) (edit) (undo)
 
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Consider a particle on the surface of the planet.
Consider a particle on the surface of the planet.
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<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
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<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}
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& {{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>

Current revision

Consider a particle on the surface of the planet.

\displaystyle F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{\, 2}}}

where \displaystyle F is the gravitational force on a particle on the surface of the planet, in other words, its weight.

\displaystyle {{m}_{1}} is the mass of the planet, \displaystyle {{m}_{2}} is the mass of the particle and \displaystyle d is the radius of the planet.

As \displaystyle F={{m}_{2}}a where \displaystyle a is the acceleration of the particle very close to the surface of the planet,


\displaystyle a=\frac{G{{m}_{1}}}{{{d}^{\, 2}}}

Thus

\displaystyle {{d}^{\, 2}}=\frac{G{{m}_{1}}}{a}

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} & {{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}

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