Please solve q 99?

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Answer 1 · 2018-04-04T09:45:04

$"(4)"\ 1.07 × 10^4\ "m/s"$

Total energy while launching the body

$"T.E"_i = -"GMm"/"R"_"e" + 1/2\ "mv"^2$

Total energy when it just reaches a height of $"10R"_"e"$

$"T.E"_f = -"GMm"/("R"_"e" + "10R"_"e") = -"GMm"/(11"R"_"e")$

$"T.E"_i = "T.E"_f$

$-"GMm"/"R"_"e" + 1/2\ "mv"^2 = -"GMm"/("11R"_"e")$

Divide both sides by $"m"$

$-"GM"/"R"_"e" + "v"^2/2= -"GM"/("11R"_"e")$

$"v"^2/2 = "GM"/"R"_"e"[1 - 1/11]$

$"v" = sqrt("20GM"/"11R"_"e")$

Substitute the values

$"v" = sqrt((20 × 6.67 × 10^-11\ "Nm"^2"/kg"^2 × 6 × 10^24\ "kg")/(11 × 6.4 × 10^6\ "m")$

$"v" ≈ 1.07 × 10^4\ "m/s"$