You are exploring a distant planet. When your spaceship is in a circular orbit a

Question

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 5500 m/s . By observing the planet, you determine its radius to be 4.48×106 m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 14.6 m/s at an angle of 30.8 above the horizontal.If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile?

Explanation / Answer

h = height above the surface of planet = 630 km = 6.30 x 105 m

r = radius of planet = 4.48 x 106 m

M = mass of planet

v = orbital speed = 5500 m/s

orbital speed is given as

v = sqrt(GM/(r + h))

5500 = sqrt((6.67 x 10-11) M/((4.48 x 106) + (6.30 x 105)))

M = 2.32 x 1024 kg

acceleration due to gravity is given as

a = GM/r2

a = (6.67 x 10-11) (2.32 x 1024)/(4.48 x 106)2

a = 7.71 m/s2

consider the motion of the projectile along the Y-direction

Voy = initial velocity = 14.6 Sin30.8 = 7.5 m/s

a = acceleration = - 7.71

Y = displacement = 0

t = time of travel

using the equation

Y = Voy t + (0.5) a t2

0 = 7.5 t + (0.5) (- 7.71) t2

t = 1.95 sec

consider the motion along the X-direction

Vox = initial velocity = 14.6 Cos30.8

a = acceleration = 0

X = displacement = ?

t = time of travel = 1.95

using the equation

X = Vox t + (0.5) a t2

X = (14.6 Cos30.8) (1.95) + (0.5) (0) (1.95)2

X = 24.5 m

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