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 5200 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

R = 4.48 x 10^6 m

radius of circular path, r = R + h = (4.48 x 10^6) + (630 x 10^3)

r = 5.11 x 10^6 m

Fg = m a_c

G M m / r^2 = m v^2 / r

G M = v^2 r

(6.67 x 10^-11) (M) = (5200^2)(5.11 x 10^6)

M = 2.072 x 10^24 kg

acceleration due to gravity, g = G M / R^2

g =6.88 m/s^2

R = v0^2 sin(2 theta) / g

Range = (14.6^2)(sin61.6) / 6.88

= 27.3 m .......Ans

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