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calculate the work required to launch a satellite from the surface of the earth (radius6380km) to...

Question:

Calculate the work required to launch a satellite from the surface of the earth (radius6380km) to orbit 6000 km above the surface.

Energy

The word which is used to show the ability of a system to execute work is represented by energy. The energy is obtained in a system in many configurations like mechanical energy, potential energy or kinetic energy, etc.

Answer and Explanation:

Given info


  • Radius of earth is {eq}{R_e} = 6380\;{\rm{km}} {/eq}
  • Radius of orbit is {eq}{R_o} = 6000\;{\rm{km}} {/eq}


The expression of kinetic energy is,

{eq}\begin{align*} K.E &= \dfrac{{m{V^2}}}{{{R_o}}}\\ \dfrac{{GMm}}{{{R_o}^2}}& = \dfrac{{m{V^2}}}{{{R_o}}}\\ V &= \sqrt {\dfrac{{GM}}{{{R_o}}}} \end{align*} {/eq}

The expression of total work done is,

{eq}\begin{align*} {\rm{Work}}\;{\rm{done}} &= P.E + K.E\\ &= \dfrac{{GMm}}{{{R_e}}} - \dfrac{{GMm}}{{{R_o}}} + \dfrac{1}{2}m{V^2}\\ &= \dfrac{{GMm}}{{{R_e}}} - \dfrac{{GMm}}{{{R_o}}} + \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{{{R_o}}}} } \right)^2}\\ &= \dfrac{{GMm}}{{{R_e}}} - \dfrac{{GMm}}{{{R_o}}} + \dfrac{1}{2}\dfrac{{GMm}}{{{R_o}}}\\ &= GMm\left( {\dfrac{1}{{{R_e}}} - \dfrac{1}{{{R_o}}} + \dfrac{1}{{2Ro}}} \right)\\ &= GMm\left( {\dfrac{1}{{{R_e}}} - \dfrac{1}{{2{R_o}}}} \right) \end{align*} {/eq}

Here, G is the gravitational constant, M is the mass of earth and m is the mass of satellite. The value of M is {eq}5.98 \times {10^{24}}\;{\rm{kg}} {/eq} and the value of G is {eq}6.67 \times {10^{ - 11}}{{{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}} {\left/ {\vphantom {{{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}} {{\rm{k}}{{\rm{g}}^{\rm{2}}}}}} \right. } {{\rm{k}}{{\rm{g}}^{\rm{2}}}}} {/eq}.


Substitute the all values.

{eq}\begin{align*} {\rm{Work}}\;{\rm{done}} &= \left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}} \right)\left( {5.98 \times {{10}^{24}}\;{\rm{kg}}} \right)m\left[ {\dfrac{1}{{\left( {6380\;{\rm{km}}} \right)}} - \dfrac{1}{{2\left( {6000\;{\rm{km}}} \right)}}} \right]\\ &= \left( {398.87 \times {{10}^{12}}\;{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/kg}}} \right)m\left[ {\dfrac{1}{{6380\;{\rm{km}}}} - \dfrac{1}{{12000\;{\rm{km}}}}} \right]\\ &= m\left( {\dfrac{{281 \times 398.87 \times {{10}^{12}}\;{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/kg}}}}{{3828000 \times {{10}^3}\;{\rm{m}}}}} \right)\;\\ &= m\left( {{\rm{29279642}}{\rm{.11}}\;{\rm{N}}{\rm{.m/kg}}} \right)\;{\rm{J}} \end{align*} {/eq}

Here, the work done in depending on the mass of the satellite.


Thus, the work done is {eq}m\left( {{\rm{29279642}}{\rm{.11}}\;{\rm{N}}{\rm{.m/kg}}} \right)\;{\rm{J}} {/eq}.


Learn more about this topic:

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What is Energy Conservation? - Definition, Process & Examples

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6
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