Question

You land a 10,000 kg spacecraft on a new planet you name Wiley. To get the lay of the land you make the following measurements: A 2.5 kg mass thrown upward at 10 m/s returns to your hand in 5 seconds. You also notice there is no atmosphere on Wiley and, therefore, no air resistance. You note that the circumference of Wiley is 2.00 x 105 km. (a) What is the mass of Wiley? (b) Wiley rotates every 1.5 Earth days. What altitude would your spacecraft need to orbit to stay fixed above a spot on Wiley’s equator (geosynchronous)? (c) How much work would your spacecraft engines need to do to reach that altitude starting from the surface of the planet? (Recall that work is equal to the change in mechanical energy. You can assume the geosynchronous orbit is circular.)

Answer 1

(a)

v = u + gt

at maximum height, v = 0 ( final velocity)

as mass returns to hand in 5 seconds, this means it took 2.5 seconds to reach max height

so,

g = 10 / 2.5

g = 4 m/s²

and

2r = 2e5 km = 2e8 m

so,

r = 3.183e7 meters

so, we know

g = GM / r²

solve for M, we have

M = gr² / G

M = 4 * 3.183e7² / 6.67e-11

M = 6.076e25 Kg

________________________________________________

(b)

T² = (4² / GM) r³

(1.5 * 24 * 3600)² = ( (4² / 6.67e-11 * 6.076e25) * r³

solve for 'r', we get

r = 1.2e8 m ( from the center of wiley)

or

r = 8.81e7 m ( above the surface of wiley)

__________________________

(c)

work done = change in P.E

P.E = - GMm / r

final P.E = - 6.67e-11 * 6.076e25 * 10000 / 1.2e8 = - 3.377e11 J

initial P.E = - 6.67e-11 * 6.076e25 * 10000 / 3.183e7 = - 1.273e12 J

so,

Work = - 3.377e11 - ( - 1.273e12)

Work = 9.355e11 J