I lied I didn't get kg^(1/2) m^(1/2) s
like I said before I prove my answer I got this which is the last line of my proof
(Kg^(1/2) s)/(m^(1/2))
Hello recently I was asked to find the time it takes a satalite that is geocentric (that always stays above the same exact spot of the earth)
so... I was given the
radius of 61 earth radius or simply 3.89 E 6 m
mass of earth
what G equals
and have to use Newtons law of univeral gravitation or this formula,
Fg = G ((m1 m2)/r^2)
ok so I rearanged the formula after setting
Fc = Fg right and rearanged for time
and got this
delta t = (2 pi r)/((G(Mc/r)^(1/2)))
ok and I got for my answer
delta t = 2.34 E 15 ...
and have to prove units which is were my problem comes along I simplified and got this
kg^(1/2) m^(1/2) s
maybe you could show me how to prove the units because I believe that I'm suppose to get seconds but I don't. Could you show me step by step how to prove the units...
here's what I did
m/((N m^2 m/(Kg^2 m))^(1/2)
m/((kg m m^2 m)/(Kg^2 m s^2))^(1/2)
m/((kg m^4)/(Kg^2 m s^2))^(1/2)
(m (Kg^2 m s^2)^(1/2))/((Kg m^4)^(1/2)
(m Kg m^(1/2) s)/(Kg^(1/2) m^2)
(m^(3/2) Kg s)/(Kg^(1/2) m^2)
m^(-1/2) Kg^(1/2) s
(Kg^(1/2) s)/(m^(1/2))
I didn't get seconds how come help me show me step by step it's driving me crazy
3 answers
I don't know what to do
The time required for an orbit is
t = 2 pi R/V
The velocity V can be obtained from
V^2/R = G M/R^2
V^2 = GM/R
V = sqrt (GM/R)
t should turn out to be 24 hours, expressed in seconds
The units of V that you get from the above formula are
sqrt [N m^2*kg/(kg^2*m)]
=sqrt [kg^2 m^2/(s^2*kg^2)]
=sqrt [m^2/s^2) = m/s
Dividing 2 pi R by V will clearly gve you units of seconds
t = 2 pi R/V
The velocity V can be obtained from
V^2/R = G M/R^2
V^2 = GM/R
V = sqrt (GM/R)
t should turn out to be 24 hours, expressed in seconds
The units of V that you get from the above formula are
sqrt [N m^2*kg/(kg^2*m)]
=sqrt [kg^2 m^2/(s^2*kg^2)]
=sqrt [m^2/s^2) = m/s
Dividing 2 pi R by V will clearly gve you units of seconds
2 answers