# position and movement

(a) A satellite orbits around the earth once every 1.5 h. The satellite is 350km

above the ground.The radius of the earth is 6370km.Find the average speed of the satellite.

ans is 2.83x10^4 km h^-1我要算式

(b) Another satellite moves in another orbit.It completes 1 revolution in 3.2 h with an average speed of 6000ms^-1 . Find the height of its orbit above the ground.

ans is 4630km 我要算式

Update:

Physics king 我用過你條式但係都係計到2.81x10^4 而ans係2.83x10^4

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(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。