You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surface, the ship's orbital speed is 4900 m/s. By observing the planet, you determine its radius to be 4.48*106m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 13.6 m/s at an angle of 30.8∘ above the horizontal. If the resistant due to plantes atmosphere is neglagible what is the horizantal range of the projectile?

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 * 10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g = GM / R² = (v² * (R+h)) / (R²)

⇒ with v = the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 * 10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * (4.48*10^6 + 630000)) / ((4.48*10^6) ²)

g = 6.11 m/s²

Step 3: Describe the position of the projectile

Horizontal component: x (t) = v0*t * cos∅

Vertical component: y (t) = v0*t * sin∅ - 1/2 gt² (will be reduced to 0 in time)

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t = v (sin∅) / g = 1.14 s

Horizontal range d = v0²/g * 2sin∅cos∅ = v0²/g * sin2∅

Horizontal range d = (13.6²) / 6.11 * sin (2*30.8)

Horizontal range d = 26.63 m

The horizontal range of the projectile = 26.63 meters