Question

A projectile is fired vertically upward from the surface of planet Moorun of mass 2 à 10^24 kg and radius 7 à 10^6 m. If this projectile is to rise to a maximum height above the surface of Moorun equal to 6 à 10^6 m, what must be the initial speed of the projectile? The universal gravitational constant is 6.67259 à 10^â11 N · m2 /kg^2 . Answer in units of km/s.

Answer 1

To solve this problem we will apply the principle of energy conservation. Here we have that the gravitational potential energy must be equal to the kinetic energy of the body. So,

$PE = KE$

$\frac{GMm}{R} = \frac{1}{2} mv^2$

Here,

m = mass of projectile

G = Gravitational Universal constant

M = Mass of the planet

R = Total height from center of mass of the planet

v = Velocity

Rearraning to find the velocity we have,

$\frac{GM}{R} = \frac{1}{2} v^2$

$v = \sqrt{2\frac{GM}{R}}$

Our values are given as,

$M = 2*10^{24} kg$

$r = 7*10^6 m$

$h = 6*10^6 m$

$R = h+r = 13*10^6m$

$G = 6.67259*10^{-11} N\cdot m^2/kg^2$

Replacing we have,

$v = \sqrt{2\frac{(6.67259*10^{-11})(2*10^{24})}{13*10^6}}$

$v = 4531.12m/s$

Therefore the initial speed of the projectile must be 4531.12m/s