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 4900 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 13.6 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? Express your answer with the appropriate units.

Explanation / Answer

To find speed of spaceship, v use,
gravitaional force = centrepetal force
G*M*m/(r+h)^2 = m*v^2/(r+h)
G*M*(r+h)^2 = v^2/(r+h)

but G*M*(r+h)^2 is nothing but acceleration due to gravity, g
so above expression becomes

G*M*(r+h)^2 = v^2/(r+h)
g = v^2/(r+h)
g = 4900^2 / (4.48*10^6 + 630000)
= 4.7 m/s^2

Now for prohectile, g = 4.7 m/s^2

Range = Vo^2 * sin (2*thetha) /g
= (13.6)^2 * sin (2*30.8) / 4.7
= 34.6 m

Answer: 34.6 m

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