parking orbit (physics)

2013-02-02 5:34 am

question: what is the period and speed of the satellites parked in the parking orbit?

MY approach:T=86400, w = 2pie/T -> v/r = 2pie/T ->v = (r)(2pie/T)

and the answer i find is 465ms^-1 but the answer is actually 3010ms^-1.I want to know why?Can anyone help please?

回答 (1)

天同

2013-02-02 6:54 am

✔ 最佳答案

You have used the wrong figure for r.

r is the distance of the stationary satellite from the centre of the earth. It can be found from the following centripetal force equation.

GmM/r^2 = mrw^2
where G is the Universal Gravitational Constant
M and m are the masses of the earth and satellite respectively
w is the angular speed of the satellite

Hence, r^3 = GM/w^2
Put in the values of G = 6.67 x 10^-11 m^3/kg.s^2, M = 5.97 x 10^24 kg and w = 2.pi/T = 7.27 x 10^-5 s^-1
the value of r is calculated to be 4.22 x 10^7 m

Hence, v = (4.22x10^7) x (2.pi/T) = 3071 m/s

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