Question

If an object projected vertically upwards with speed$,$ half the escape speed of earth$,$ the maximum height attained by it is $($R is radius of the earth$)?$

Answer 1

We know that the escape velocity of Earth $=Ve=\sqrt{\left[2gR\right]}$

$g=$ gravity due to Earth

$1/2m\left(Ve\right)²-GMm/R=0$ for the object to escape from Earth$.$

$(or,KE+PE=0)$

When the body is projected from Earth's surface$:$

Given $u=$ initial velocity $=Ve/2=\surd \left(gR/2\right)$

$InitialKE=1/2mu²=mgR/4=GMm/\left(4R\right)$

Initial $PE=-GMm/R$

total initial energy $=-3GMm/\left(4R\right)$

When the body reaches a height h above surface of Earth and stops$:$

$v=0.$ Final $KE=0.$

Final $PE=-GMm/(R+h)$

From conservation of Energy$:$

$-GMm/(R+h)=-3Gm/\left(4R\right)$

$3(R+h)=4R$

$h=R/3$

$AnswerisR/3.$