position and movement

(a) Distance covered by satellite in one revolution round the earth
= 2 x pi x (350+6370) km
where pi = 3.14159.....

Hence, average speed = distance/time = 2 x pi x (350+6370) /1.5 km/h

(b) Similar, distance covered by satellite in one revolution
= 2 x pi x (h+ 6370) km
where h (in km) is the height of the satellite above earth surface

hence, 2 x pi x (h+ 6370) = (6000x3.6) x 3.2
solve for h

a. Radius of orbit of the satellite, R = 6370 + 350 = 6720 km = 6.720 X 10^6 m



Period, T = 1.5 h = 1.5 X 3600 = 5400 s



By v = 2piR / T



v = 2pi(6.720 X 10^6) / 5400



Speed, v = 7.82 X 10^3 m/s



v = 2.83 X 10^4 km/h


b. By 2piR / v = T

2piR / 6000 = 3.2 X 3600

Radius of orbit, R = 1.10 X 10^7 m

So, the height from the ground = 1.10 X 10^7 - 6.37 X 10^6 = 4.63 X 10^6 m

= 4630 km


2009-10-19 21:25:09 補充：
這麼小的誤差，在物理學上是不值一提的。

2009-10-19 21:26:01 補充：
只要你的方法是對的，便可以了。

可能modal answer是在計算中進了位，所以便有誤差。
我們是將計數機的小數點一直保留至答案，應該是我們對，modal answer錯!!!