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 June 17th, 2013, 09:24 PM #1 Newbie   Joined: Jun 2013 Posts: 13 Thanks: 0 Acceleration Due to Gravity Determine the value of g on earth, the radius of the earth is given by 6000 Km and the density 7.5 g/cc. The value of $G= 6.66*10^{-10}Nm^{2}/Kg^{2}$
 June 17th, 2013, 10:17 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Acceleration Due to Gravity Use Newton's law of gravity: $F=G\frac{mM}{r^2}$ Now, on the left side use Newton's 2nd law of motion: $mg=G\frac{mM}{r^2}$ Divide through by $m$: $g=G\frac{M}{r^2}$ Now, what you need to do is determine the mass of the Earth, using the radius (assume it is a sphere) and the given mass density): $M=\rho V=\rho \frac{4}{3}\pi r^3$ and we now have: $g=G\frac{\rho \frac{4}{3}\pi r^3}{r^2}=G\rho \frac{4}{3}\pi r$ Next, convert $\rho$ to kg/m³: $\rho=7.5\text{ \frac{g}{cc}}\cdot\text{\frac{1 kg}{1000 g}}\cdot$$\text{\frac{100 cm}{1 m}}$$^3=7500\text{ \frac{kg}{m^3}}$ Now, plugging in the data (your value of G is ten times too large), we find: $g=$$6.66\times10^{-11}$$(7500)\frac{4}{3}\pi(6000000)\approx12.6\text { \frac{m}{s^2}}$ If we use more accurate data, we find: $g=$$6.673848\times10^{-11}$$(5520)\frac{4}{3}\pi(6371000)\approx9.8\text{ \frac{m}{s^2}}$

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