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October 1, 2014

October 1, 2014

Posted by **Physics** on Thursday, May 7, 2009 at 9:01pm.

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

- Physics -
**Physics**, Thursday, May 7, 2009 at 9:02pmI 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))

- HELP ME! -
**HELP ME!**, Thursday, May 7, 2009 at 9:08pmI don't know what to do

- Physics -
**drwls**, Thursday, May 7, 2009 at 9:29pmThe 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

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