Question

A projectile is shot straight up from the earth’s surface at a speed of $10,000km/h$. How high does it go?

Short answer

Height of the projectile$335665.44m$

Step-by-step solution

Given information

speed of the projectile$v=10,000km/h$

Gravitational constant $G=6.673\times {10}^{-11}N{m}^2/k{g}^2$,

Mass of the earth $M=5.9742\times {10}^{24}kg$

Radius of earth$r=6378100m$

Explanation

Naturally we assume that $10000km/hr$is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. $M$is earth mass,$r$is earth radius.

$$\begin{gathered} \frac{KE}{m}=\frac{1}{2}{\left(\frac{10000\times 1000}{3600}\right)}^2=3858025J/kg \\ \frac{∆PE}{m}=\frac{P{E}_{atheight}-P{E}_{atsurface}}{m}=-\frac{GM}{(r+h)}+\frac{GM}{r} \end{gathered}$$

On net energy in equilirbium,

$$\begin{gathered} \frac{KE}{m}=\frac{∆PE}{m} \\ \frac{KE}{m}-\frac{GM}{r}=-\frac{GM}{(r+h)} \\ h=-\frac{GM}{\left({\displaystyle \frac{KE}{m}}-{\displaystyle \frac{GM}{r}}\right)-r}=335665.44m \end{gathered}$$