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?

Answer 1

An object is thrown up with an initial velocity is v_i, So Total energy of the object is

E_i = `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, E_f = `1/2"Mv"_∞`

E_i = E_f ⇒ v_i = 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"`

v_e = `sqrt("v"_∞^2 + 2"gR"_"E")`