Show that the satellite revolving around the earth will depart forever if its speed is increased by 41.42%.

Solution

Let a satellite of mass m be revolving in circular orbit of radius 'r' around the earth of mass M.
The orbital velocity of satellite is given by,

straight v subscript straight o equals square root of GM over straight r end root

The velocity required by the satellite is obtained from the equation,

1 half mv subscript straight e squared space equals space GMm over straight r

rightwards double arrow

straight v subscript straight e equals square root of fraction numerator 2 GM over denominator straight r end fraction end root

To escape, the percentage increase in velocity is,

fraction numerator straight v subscript straight e minus straight v subscript straight o over denominator straight v subscript straight o end fraction cross times 100 space equals space fraction numerator square root of begin display style fraction numerator 2 GM over denominator straight r end fraction end style end root space minus space square root of begin display style GM over straight r end style end root over denominator square root of begin display style GM over straight r end style end root end fraction cross times 100

space space equals left parenthesis square root of 2 minus 1 right parenthesis cross times 100 space space space equals space 41.42 percent sign

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