
Solutions:
1) v = sqrt(2GM/r) = sqrt[(2)(6.67x1011 Nm2/kg2)(5.98x1024 kg)/(6.37x106m)] = 11.2 km/s
2) a = F/m = (800,000 N)/(10,000 kg) = 80 m/s2
3) 1 g = 9.81 m/s2; (80 m/s2)(1 g)/(9.81 m/s2) = 8.15 g's
4) v = at; t = v/a = (11.2 km/s)/(80 m/s) = 140 s = 2.33 min.
5) t = d/v = (10,000 ltyrs)/(0.8 ltyrs/yr) = 12,500 years
6) [delta]to = ([delta]t)sqrt((1(v/c)2)) = (10,000 years)/sqrt(0.36) = 6000 yrs
7a) V = (4[pi]/3)r3 = (4[pi]/3)(1x104m)3 = 4.19x1012m3 density = m/v = (2.0x1030kg)/(4.189x1012m3) = 4.77x1017 kg/m3
7b) (1kg)(m3/4.77x1017 kg) = (2.09x1018m3)(1x106cm3/1m3) = 2.09x1012cm3
8) Fc=FG; mv2/r = GMm/r2 => v = sqrt(GM/r) = sqrt((6.67x1011 Nm2/kg2)(2.0x1030kg)/10,000m) = 1.15x108 m/s
9) T = 2[pi]r/v = 2[pi](10,000 m)/1.15x108 m/s = 5.4104 s
10) v = sqrt(2GM/r) = sqrt((2)(6.67x1011 Nm2/kg2)(2.0x1030kg)/10,000m) = 1.63x108m/s
11) a = F/m = 50,000 N/100 kg = 500 m/s2; v=at = (500 m/s2)(10 hrs)(3600 s/hr) = 1.8x107m/s; no, not enough speed before fuel runs out.
12) Bad plan: (1) Incredibly high velocity orbit required (2) Incredibly small volume sample required (3) Thrust insufficient to escape star
13) vesc = sqrt(2GM/r) = sqrt((2)(6.67x1011 Nm2/kg2)(2.2x1030kg)/3000m) = 3.12x108 m/s; Impossible escape velocity because it is greater than the speed of light and nothing can attain such a velocity.
14 a) The bottom portion of the craft will experience the most intense gravitational field because it is closest to the black hole.
14 b) FG = GmM(1/r21/(r+10)2) = (6.67x1011 Nm2/kg2)(100kg)(2.2x1030kg)((1/1.0x108m)2 (1/1.0000001x108m)2) = 0.293 N
14c) No, the difference in gravitational force is less than 1000N, so the probe will remain intact.
14d) 6.6455E+06 m