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.
(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.
3 answers
(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
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
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]
At 38,268km,
Vc = sqrt[3.9863x10^14/6x6,378,000] =
The period derives from
T = 2(Pi)sqrt[r^3/µ]
1 answer