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g the total mechanical energy of the satellite-Earth system when the satellite is in its current orbit is E. In order for the satellite to orbit Earth in a new stable circular orbit at an altitude of 12RE, the energy of the satellite-Earth system must be

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Cricetus

Answer:

The correct answer is "[tex]\frac{4E}{3}[/tex]".

Explanation:

According to the question,

Energy of satellite,

⇒ [tex]E_s=-\frac{GM_sM_E}{2r}[/tex]

For the very 1st case:

[tex]r = R_E+R_E[/tex]

  [tex]=2R_E[/tex]

or,

⇒ [tex]E=-\frac{GM_sM_E}{4R_E}[/tex]...(1)

For the new case:

[tex]r = R_E+\frac{R_E}{2}[/tex]

  [tex]=\frac{3R_E}{2}[/tex]

then,

⇒ [tex]E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }[/tex]

        [tex]=-\frac{GM_sM_E}{3R_E}[/tex]...(2)

From equation (1) and (2), we get

⇒ [tex]E'=\frac{1}{3}(4E)[/tex]

        [tex]=\frac{4E}{3}[/tex]  

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