$$F=mg$$ Why is the acceleration constant? Shouldn't it change as it is thrown upward as the distance from the earth increases.I know the effects would be very negligible but is there any equation to account for this change? How can we apply it to real life? $$\frac{dg}{dr} =\lim_{(h \to 0)} \Bigr( \frac{Gm_{earth}}{(r+h)^2}-\frac{Gm_{earth}}{r^2}\Bigr ) \frac{1}{h}=\frac{-2Gm_{earth}}{r^3} $$

$$F=mg$$ Why is the acceleration constant? Shouldn't it change as it is thrown upward as the distance from the earth increases.I know the effects would be very negligible but is there any equation to account for this change? How can we apply it to real life? $$\frac{dg}{dr} =\lim_{h \to 0} \left[ \frac{Gm_{earth}}{(r+h)^2}-\frac{Gm_{earth}}{r^2}\right] \frac{1}{h}=\frac{-2Gm_{earth}}{r^3} $$

$$F=mg$$ Why is the acceleration constant? Shouldn't it change as it is thrown upward as the distance from the earth increases.I know the effects would be very negligible but is there any equation to account for this change? How can we apply it to real life? $$\frac{dg}{dr} =\lim(h =0) \Bigr( \frac{Gm_{earth}}{(r+h)^2}-\frac{Gm_{earth}}{r^2}\Bigr ) \frac{1}{h}=\frac{-2Gm_{earth}}{r^3} $$

$$F=mg$$ Why is the acceleration constant? Shouldn't it change as it is thrown upward as the distance from the earth increases.I know the effects would be very negligible but is there any equation to account for this change? How can we apply it to real life? $$\frac{dg}{dr} =\lim_{(h \to 0)} \Bigr( \frac{Gm_{earth}}{(r+h)^2}-\frac{Gm_{earth}}{r^2}\Bigr ) \frac{1}{h}=\frac{-2Gm_{earth}}{r^3} $$

$$F=mg$$ Why is the acceleration constant  ? Shouldn't it change as it is thrown upward as the distance from the earth increases.I know the effects would be very negligible but is there any equation to account for this change  ?How can we apply it to real life? $$\frac{dg}{dr} =\lim(h =0) \Bigr( \frac{Gm_{earth}}{(r+h)^2}-\frac{Gm_{earth}}{r^2}\Bigr ) \frac{1}{h}=\frac{-2Gm_{earth}}{r^3} $$

$$F=mg$$ Why is the acceleration constant? Shouldn't it change as it is thrown upward as the distance from the earth increases.I know the effects would be very negligible but is there any equation to account for this change? How can we apply it to real life? $$\frac{dg}{dr} =\lim(h =0) \Bigr( \frac{Gm_{earth}}{(r+h)^2}-\frac{Gm_{earth}}{r^2}\Bigr ) \frac{1}{h}=\frac{-2Gm_{earth}}{r^3} $$