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 calculate the initial speed of the projectile, we can use the concept of gravitational potential energy.

Step-by-step explanation:

To calculate the initial speed of the projectile, we can use the concept of gravitational potential energy. At the highest point of the projectile's trajectory, all of its initial kinetic energy is converted into gravitational potential energy. The equation for gravitational potential energy is given by:

P.E. = mgh

Where m is the mass of the projectile, g is the acceleration due to gravity, and h is the height. Rearranging the equation, we get:

v = sqrt((2gh))

Substituting the given values, we have:

v = sqrt((2 * 6.67259 * 10^(-11) * 2 * 10^24) / (7 * 10^6 + 6 * 10^6)) = 2.546 km/s

Answer 2

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`

`(GMm)/(R) = (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,

`(GM)/(R) = (1)/(2) v^2`

`v = \sqrt{2(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((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