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April 16, 2014

April 16, 2014

Posted by **physics** on Saturday, September 12, 2009 at 4:34pm.

ok aparently there\'s an easier way to do this

I applied Newtons second law in the radial direction

net force radial = m (radial acceleration) = Fg

were Fg is r^-2 G m m2

I solved for the period then calculated the acceleration to be

5.91 E 18 s^-3 m

how do I do this because I just assumed that it was the Earth\'s moon which... isn\'t necissarily true and i used the mass of the earth which wasn\'t given in the problem i used 5.98 E 24 kg for mass of earth

what is the easy way to do this...

my teacher told me

Little g is the force of gravity (weight) divided by the mass of the object that’s feeling the force. Another way to look at it: little g helps find the force a single kilogram of mass would feel. It describes the force per unit mass. You could say that g = W/m (which makes sense, when you consider that W = mg).

not exactly sure what the heck it\'s talking about because Fg or in that case weight is not equal to mg

- physics -
**bobpursley**, Saturday, September 12, 2009 at 4:51pmacceleartiononmoon= GMassmoon/radius^2

=G 7.35E22/(1.74E6)^2= G 2.43E10=

= 6.67E-11*2.43E10=

1.62 m/s^2 or as a gravitational field constant= 1.62 N/kg

I have no idea why you used the mass of Earth.

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