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Question
An object is thrown from Earth in such a way that it reaches a point at infinity with non-zero kinetic energy `["K"."E"("r" = ∞) = 1/2 "Mv"_"∞"^2]`, with what velocity should the object be thrown from Earth?
Solution
An object is thrown up with an initial velocity is vi, So Total energy of the object is
Ei = `1/2"Mv"_"i"^2 - "GMm"_"e"/"R"_"E"`
Now, the object reaches a height with a non-zero K.E.
K.E becomes infinity. P.E becomes zero.
So, Ef = `1/2"Mv"_∞`
Ei = Ef ⇒ vi = V∞
`1/2 "Mv"_∞^2 = 1/2"Mv"_"e"^2 - (2"GMM"_"E")/"R"_"E"`
`1/2 "Mv"_∞^2 = 1/2"M" ["V"_"e"^2 - (2"GM"_"E")/"R"_"E"]`
`"v"_"e"^2 = "v"_∞^2 + 2("GM"_"E"/"R"_"E") "R"_"E"`
= `"v"_∞^2 + 2"gR"_"E"`
ve = `sqrt("v"_∞^2 + 2"gR"_"E")`
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