# Chapter 4 - Worked Solutions

### to selected extension questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

Question 49

t2 (above the surface) = 2t1 (on surface)

s1 = ut1 + ½ g1t12

s1 = ut2 + ½ g2t22

Because u = 0 and because both sides are equal to s1 then g1t12 = g2t22

9.8 t12 = g2 t22

9.8 t12 = g2 (2t1)2

g2 = 2.45 ms-2

Alternatively: on the surface, the time to fall 1.5 m from rest can be calculated from:

s1 = ut1 + ½ g1t12

1.5 = 0 + ½ x -9.8 t12

t1 = 0.553 s

As t2 = 2 t1 then t2 = 1.107 s

s1 = ut2 + ½ g2t22

1.5 = 0 + ½ x g2 (1.107)2

g2 = 2.45 s

g = Gm/d2

d2 = Gm/g = 6.67 x 10-11 x 6 x 1024/2.45 = 1.6 x 1014

d = 1.28 x 107 m

Distance above Earth = 1.28 x 107 m - 6.38 x 106 = 6400 km

Question 50

The relationship to consider is: g µ m/d2

(a) half the mass but the same radius (or twice mass and twice radius etc).

(b) if t is twice normal then g would be ¼ normal, therefore: double the radius; keep mass constant.

(c) same as for answer (b)

Question 51

g(surface) = Gm/d2 = Gm/(5.1 x 106)2 = Gm/(2.6 x 1013)

g(above surface) = Gm/d2 = Gm/(5.1 x 106 + 150 x 103)2 = Gm/(2.76 x 1013)

g(above surface)/g(surface) = 2.6 x 1013/2.76 x 1013 = 0.94

Question 52
 r3 T2 7.5 x 1025 3.13 30.2 x 1025 12.60 122.5 x 1025 51.27 664.4 x 1025 278.89 Graph is linear (passes through origin) hence r3 µ T2

Question 53

(a) FMoon = FEarth

G mem/d2 = Gmmm/(3.8 x 108 - d)2

6 x 1024/d2 = 7.34 x 1022/(3.8 x 108 - d)2

Take square roots of both sides:

2.45 x 1012/d = 2.71 x 1011/(3.8 x 108 -d)

9.31 x 1020 = 2.45 x 1012 d = 2.71 x 1011 d

9.31 x 1020 = 2.72 x 1012 d

d = 3.42 x 108 m

(b) 0.89 of the distance to the Moon

(c) NO! In circular orbit you would experience free-fall and feel weightless.

Question 54

(a) g = Gm/d2 = 6.67 x 10-11 x 6 x 1024/(6.38 x 106 + 10000)2 = 9.8011 ms-2

No noticable difference.

(b) Same as on the Earth's surface - no difference.

Question 55

(a) g = Gm/d2 = 6.67 x 10-11 x 2.0 x 1030/(12 x 103)2 = 9.2 x 1011 ms-2

(b) v = 2p r/T = 2 p x 12 x 103/0.041 = 1.84 x 106 ms-1

ac = v2/r = (1.84 x 106)2/ 1.2 x 103 = 2.8 x 108 ms-2

Question 56

Spring clock would not be affected by gravity and therefore will keep time. A pendulum is affected by gravity and won't keep time. The greater the gravitational force the faster the pendulum will oscillate.

Question 57 re = 6.38 x 106 m

rb (radius for Brisbane) = re cos 28° = 6.38 x 106 x cos 28° = 5.63 x 106 m

(a) circumference = 2 p r = 2 p x 5.63 x 106 = 3.54 x 107 m

(b) v = s/t = 3.54 x 107/ (24 x 3600) = 410 ms-1

(c) ac = v2/r = 4102/5.63 x 106 = 0.030 ms-2

(d) FC = mac = 60 x 0.030 = 1.79 N

(e) (i) Gravitational force FG points towards the centre of the Earth (see figure that follows): (ii) Perp. to Earth's axis through Brisbane (see figure that follows): (iii) Perp. to Earth's axis through Brisbane (see figure for (ii) above).

Question 58

v = √(2Gm/r) = √(2 x 6.67 x 10-11 x 1.4 x 1011 x 2 x 1030)/(80000 x 3600 x 24 x 365 x 3 x 108)
= (3.73 x 1031)/(7.57 x 1020) = 4.93 x 1010 m/s (greater than "c" so can't escape).

Question 59

r = 9.4 x 106m
T = 7h 39m = 7.65h = 2.754 x 104 s
v = √(2pr/T) = (5.9 x 107/(2.754 x 104) = 2.14 x 103 m/s

Fc= mv2/r = Gm1m2/r2
v2 = Gm/r
m = v2r/G = (2.14 x 103)2 x 9.4 x 106/(6.67 x 10-11)
m = 6.475 x 1023 kg

Question 60

(a) lp = 0.0029/T = 0.0029/7600 = 3.8 x 10-7 m
lf = 0.0029/T = 0.0029/7600 = 3.8 x 10-7 m
(b) Procycon as it has bigger surface area.
(c) Same temp, same area, hence same rate of emission.