Question

A projectile is fired straight upward from the Earth's surface at the South Pole with an initial speed equal to one third the escape speed. (The radius of the Earth is 6.38 106 m.)
(a) Ignoring air resistance, determine how far from the center of the Earth the projectile travels before stopping momentarily. m
(b) What is the altitude of the projectile at this instant? m

Answer 1

Answer:

7177500 m

797500 m

Explanation:

G = Gravitational constant

m = Mass of projectile

M = Mass of Earth

R = Radius of Earth

h = Altitude

r = Distance from the center of Earth

$v_e$ = Escape velocity

Initial velocity

$u=\frac{1}{3}v_e\\\Rightarrow u=\frac{1}{3}\sqrt{\frac{2GM}{R}}$

Kinetic energy at the surface

$K=\frac{1}{2}mu^2\\\Rightarrow K=\frac{1}{2}m\left(\frac{1}{3}\sqrt{\frac{2GM}{R}}\right)^2\\\Rightarrow K=\frac{1}{9}m\frac{GM}{R}$

Potential + Kinetic energy at the surface = Potential energy at the max height

$-\frac{GMm}{R}+\frac{1}{9}m\frac{GM}{R}=\frac{GMm}{r}$

Cancelling G, M, and m

$-\frac{1}{R}+\frac{1}{9}\frac{1}{R}=-\frac{1}{r}\\\Rightarrow \frac{-9+1}{9R}=-\frac{1}{r}\\\Rightarrow \frac{8}{9R}=\frac{1}{r}\\\Rightarrow r=\frac{9}{8}R\\\Rightarrow r=\frac{9}{8}\times 6.38\times 10^6\\\Rightarrow r=7177500\ m$

Distance of the max height from the center of earth is 7177500 m

$R+h = r\\\Rightarrow h=r-R\\\Rightarrow h=7177500-6.38\times 10^6\\\Rightarrow h=797500\ m$

The altitude of the projectile is 797500 m