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Solutions:

1) v = sqrt(2GM/r) = sqrt[(2)(6.67x10-11 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 lt-yrs)/(0.8 lt-yrs/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) = (2.09x10-18m3)(1x106cm3/1m3) = 2.09x10-12cm3

8) Fc=FG; mv2/r = GMm/r2 => v = sqrt(GM/r) = sqrt((6.67x10-11 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 = 5.410-4 s

10) v = sqrt(2GM/r) = sqrt((2)(6.67x10-11 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)(6.67x10-11 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/r2-1/(r+10)2) = (6.67x10-11 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

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