Solution 2.6

From Mechanics

Revision as of 14:23, 12 March 2010

Use the theory of gravitation equation

\displaystyle F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}}

where,

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

In this case

\displaystyle {{m}_{1}} is the mass of the earth and \displaystyle {{m}_{2}} is the mass of the satellite.

\displaystyle d 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

\displaystyle {{m}_{1}}= \displaystyle \text{5}\textrm{.}\text{98}\times \text{1}0^{\text{24}}\text{ } kg

\displaystyle {{m}_{2}}=300\ kg

The radius of the earth is

\displaystyle \text{6}\textrm{.}\text{37}\times \text{1}0^{\text{6}}\text{ m }

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

which gives

\displaystyle d=\displaystyle \text{13}\textrm{.}\text{37}\times \text{1}0^{\text{6}}\text{ m }

Inserting all these quantities in the gravitation equation

\displaystyle \begin{align} & 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\textrm{.}66\times {{10}^{15}}}{178\textrm{.}76\times {{10}^{12}}}=0\textrm{.}669\times {{10}^{3}}\ =669\ \text{N} \\ \end{align}