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

given orbital speed of spaceship, vo = 4900 m/s

h = 630 km

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

Let M is the mass of planet.

radius of orbit of spaceship, r = R + h

= 4.48*10^6 + 630*10^3

= 5.11*10^6 m

we know, Orbital speed, vo = sqrt(G*M/r)

vo^2 = G*M/r

==> M = vo^2*r/G

= 4900^2*5.11*10^6/(6.67*10^-11)

= 1.84*10^24 kg

so, acceleleration due to gravity on the planet, g = G*M/R^2

= 6.67*10^-11*1.84*10^24/(4.48*10^6)^2

= 6.11 m/s^2

given projectile inituila speed, v = 13.6 m/s

angle of projectuon, theta = 30.8 degrees

we horizontal range, R = v^2*sin(2*theta)/g

= 13.6^2*sin(2*30.8)/6.11

= 26.63 m <<<<<<<<<<<<<<<-------------------------Answer

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