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Solution 2.8

From Mechanics

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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)
 
Line 1: Line 1:
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
Line 21: Line 21:
-
<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
Line 33: Line 33:
<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>

Current revision

Consider a particle on the surface of the planet.

F=d2Gm1m2

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

m1 is the mass of the planet, m2 is the mass of the particle and d is the radius of the planet.

As F=m2a where a is the acceleration of the particle very close to the surface of the planet,


a=d2Gm1

Thus

d2=aGm1

As G=6.6710−11 kg-1m3s-2 we get


d2=3.26.6710−1151020=3.233.35109=10.4109=1.041010d=1.04105=1.02105 m=102 km

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