Question

Please solve q 99?

Answer 1

`"(4)"\ 1.07 × 10^4\ "m/s"`

Explanation:

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")`

By conservation of energy

`"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"`