Posted by **Anonymous** on Thursday, March 10, 2011 at 6:48pm.

(a) Find the speed of a satellite moving around the earth in a circular orbit that has a radius equal to six times the earth's radius of 6.38 x 10^6 m.

(b) Find the satellite's orbital period.

- Physics -
**bobpursley**, Thursday, March 10, 2011 at 6:58pm
centripetalforce=gravity force

v^2/r=9.8/(6)^2

V= sqrt (6*re*9.8/36)

checkthat.

Period? Well, T=distance/velocity=2PI*6Re/above V

That will give the period in seconds, you probably want to convert it.

- Physics -
**drwls**, Thursday, March 10, 2011 at 7:04pm
(a) centrieptal acceleration rate =

V^2/r = GM/r^2

where M is the mass of the earth and r = 6 Re.

Re is the radius of the earth that you were given

G is the universal constant of gravity.

Solve for V

V^2/(6 Re) = GM/(6Re)^2

You can save yourself the trouble of looking up M and G by using the relationship

g = GM/Re^2. Then

6 V^2 /Re = GM/Re^2 = g

V^2 = (Re*g/6)

V = 3230 m/s

(b) V*(period) = 2 pi R = 12 pi Re

Solve for the period

- Physics -
**tchrwill**, Friday, March 11, 2011 at 10:26am
Vc = sqrt(µ/r) where µ = 3.9863x10^14 and r = 6378km.

At 38,268km,

Vc = sqrt[3.9863x10^14/6x6,378,000] =

The period derives from

T = 2(Pi)sqrt[r^3/µ]

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