Question

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed v0(v < vesc). Express in terms of v0, rE, ME, and G.

Answer 1

Initially, the energies are:

`U_(i)=-(G M_(\varepsilon) m)/(r_(e)) \\ =K_(i)=(1)/(2) m v_(0)^(2)`

At final point, the energies are:

`U_(f)=-(G M_(\varepsilon) m)/(r_(e)+h) \\ K_(f)=(1)/(2) m(0)^(2)=0`

Using conservation law of energy,

`-(G M_(e) m)/(r_(e))+(1)/(2) m v_(0)^(2) &=-(G M_(e) m)/(r_(\varepsilon)+h) \\ -(G M_(e))/(r_(e))+(v_(0)^(2))/(2) &=-(G M_(e))/(r_(e)+h) \\ (-2 G M_(e)+r_(e) v_(0)^(2))/(2 r_(e)) &=-(G M_(e))/(r_(e)+h) \\ (r_(e)+h)/(G M_(e)) &=(2 r_(e))/(2 G M_(e)-r_(e) v_(0)^(2))`

The equation is further simplified as:

`r_(e)+h &=\left((2 r_(e))/(2 G M_(e)-r_(e) v_(0)^(2))\right) G M_(e) \\ h &=(2 r_(e) G M_(e))/(2 G M_(e)-r_(e) v_(0)^(2))-r_(e) \\ &=(2 r_(e) G M_(e)-2 r_(e) G M_(e)+r_(e)^(2) v_(0)^(2))/(2 G M_(e)-r_(e) v_(0)^(2)) \\ & h=(r_(e)^(2) v_(0)^(2))/(2 G M_(e)-r_(e) v_(0)^(2))`