t_{2}(above the surface) = 2t_{1}(on surface)

s_{1}= ut_{1}+ ½ g_{1}t_{1}^{2 }

s_{1}= ut_{2}+ ½ g_{2}t_{2}^{2 }

Because u = 0 and because both sides are equal to s_{1}then g_{1}t_{1}^{2 }_{= }g_{2}t_{2}^{2 }

9.8 t_{1}^{2}_{ = }g_{2}t_{2}^{2 }

9.8 t_{1}^{2}_{ = }g_{2}(2t_{1})^{2 }

g_{2}= 2.45 ms^{-2 }

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

s_{1}= ut_{1}+ ½ g_{1}t_{1}^{2 }

1.5 = 0 + ½ x^{-}9.8 t_{1}^{2 }

t_{1}= 0.553 s

As t_{2}= 2 t_{1 }then t_{2}=_{ }1.107 s

s_{1}= ut_{2}+ ½ g_{2}t_{2}^{2 }

1.5 = 0 + ½ x g_{2}(1.107)^{2 }

g_{2}= 2.45 s

g = Gm/d^{2 }

d^{2}= Gm/g = 6.67 x 10^{-11}x 6 x 10^{24}/2.45 = 1.6 x 10^{14 }

d = 1.28 x 10^{7}m

Distance above Earth = 1.28 x 10^{7}m - 6.38 x 10^{6}= 6400 km

Question 50

The relationship to consider is: g µ m/d^{2 }

(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)

g_{(surface)}= Gm/d^{2}= Gm/(5.1 x 10^{6})^{2}= Gm/(2.6 x 10^{13})

g_{(above surface)}= Gm/d^{2}= Gm/(5.1 x 10^{6}_{ }+ 150 x 10^{3})^{2}= Gm/(2.76 x 10^{13})

g_{(above surface)}/g_{(surface)}= 2.6 x 10^{13}/2.76 x 10^{13}= 0.94

r

^{3}T

^{2}7.5 x 10

^{25}3.13

30.2 x 10

^{25}12.60

122.5 x 10

^{25}51.27

664.4 x 10

^{25}278.89

Graph is linear (passes through origin) hence r^{3}µ T^{2 }

Question 53

(a)

F_{Moon}= F_{Earth}

G m_{e}m/d^{2}= Gm_{m}m/(3.8 x 10^{8}- d)^{2 }

6 x 10^{24}/d^{2}= 7.34 x 10^{22}/(3.8 x 10^{8}- d)^{2 }

Take square roots of both sides:

2.45 x 10^{12}/d = 2.71 x 10^{11}/(3.8 x 10^{8}-d)

9.31 x 10^{20}= 2.45 x 10^{12}d = 2.71 x 10^{11}d

9.31 x 10^{20}= 2.72 x 10^{12}d

d = 3.42 x 10^{8}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/d^{2}= 6.67 x 10^{-11}x 6 x 10^{24}/(6.38 x 10^{6}+ 10000)^{2}= 9.8011 ms^{-2 }

No noticable difference.

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

Question 55

(a) g = Gm/d^{2}= 6.67 x 10^{-11}x 2.0 x 10^{30}/(12 x 10^{3})^{2}= 9.2 x 10^{11}ms^{-2 }

(b) v = 2p r/T = 2 p x 12 x 10^{3}/0.041 = 1.84 x 10^{6}ms^{-1 }

a_{c}= v^{2}/r = (1.84 x 10^{6})^{2}/ 1.2 x 10^{3}= 2.8 x 10^{8}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

r_{e}= 6.38 x 10^{6}m

r_{b}(radius for Brisbane) = r_{e}cos 28° = 6.38 x 10^{6}x cos 28° = 5.63 x 10^{6}m

(a) circumference = 2 p r = 2 p x 5.63 x 10^{6}= 3.54 x 10^{7}m

(b) v = s/t = 3.54 x 10^{7}/ (24 x 3600) = 410 ms^{-1 }

(c) a_{c}= v^{2}/r = 410^{2}/5.63 x 10^{6}= 0.030 ms^{-2 }

(d) F_{C}= ma_{c}= 60 x 0.030 = 1.79 N

(e) (i) Gravitational force F_{G}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 10^{11}x 2 x 10^{30})/(80000 x 3600 x 24 x 365 x 3 x 10^{8})

= (3.73 x 10^{31})/(7.57 x 10^{20}) = 4.93 x 10^{10}m/s (greater than "c" so can't escape).

**Question 59**

r = 9.4 x 10

^{6}m

T = 7h 39m = 7.65h = 2.754 x 10^{4}s

v = √(2pr/T) = (5.9 x 10^{7}/(2.754 x 10^{4}) = 2.14 x 10^{3}m/s

F_{c}= mv^{2}/r = Gm_{1}m_{2}/r^{2 }v^{2}= Gm/r

m = v^{2}r/G = (2.14 x 10^{3})^{2}x 9.4 x 10^{6}/(6.67 x 10^{-11})

m = 6.475 x 10^{23}kg

**Question 60**

(a) l

_{p}= 0.0029/T = 0.0029/7600 = 3.8 x 10^{-7}m

l_{f}= 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.

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