why we use g=9.81m/s^s near area of earth?

Question

anonymous · Answer

The following video at 3:00 begins the derivation. I hope it helps. I can' t really explain it. http://www.khanacademy.org/science/physics/mechanics/v/mass-and-weight-clarification

shane_b · Answer

$$\large F_g=G\frac{m_{earth}}{r_{earth}^2}$$$$\large G (\text{gravitational constant})=6.673*10^{-11} m^3 kg^{-1}s^{-2}$$$$\large m_{earth}=5.9742x10^{24}kg$$$$\large r_{earth}=6378.1km$$$$\large F_g=(6.673*10^{-11} m^3 kg^{-1}s^{-2})\frac{5.9742x10^{24}kg}{(6,378,100m)^2}=\text{~}9.8m/s^2$$

ghazi · Answer

@shane_B if i am not mistaken, i guess unit of Force is N , secondly $$F=G \frac{ m1*m2 }{ r^2 }$$ force of attraction between two heavenly bodies

ghazi · Answer

$$a= \frac{ F }{ m1 }= G \frac{ M _{earth} }{ R^2 _{earth} }$$

ghazi · Answer

$$M _{2}=M _{earth}$$

ghazi · Answer

@vannayen

shane_b · Answer

@ghazi: You're right...I oversimplified it and should have included the second mass in the calculation.  I basically assumed a mass of 1kg for m2 and left it out of the equation.  The answer should have been in kg*m/s^2 (N) which when divided by 1kg = 9.8m/s^2.

@demitris: Nice image and good point about the radius varying between the equator and the poles.

@vannayen: Note that this is not how we actually *measure* the force of gravity.  That can be done using several other methods (atwood machine, springs, etc).