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When presented with a problem asking for the orbital velocity of an object $1,000 km$ above the Earth's surface, the formula I know you can use is $$v=\sqrt{\frac{G*m}{r}}$$ However, I was thinking that since we know the acceleration due to Earth's gravity (and because it doesn't diminish very much in orbit), then we may be able to use the formula $$v=\sqrt{g*r}$$

The reason I think this might work comes from the derivation of the first formula. $$\frac{v^2}{r}=\frac{G*m}{r^2}$$ Multiplying both sides by $r$ gives $$v^2=\frac{G*m}{r^2}*r$$ Since $\frac{G*m}{r^2}$ is the acceleration due to gravity, we can replace it with $g=9.81 m/s^2$, and thus get the formula $$v^2=g*r$$ Solving for $v$ results in $$v=\sqrt{g*r}$$

So, would this work? I tried it on a few problems, but I couldn't get a matching answer (it was less than 1 unit off). I'm going to revisit those problems to see if I made a mistake typing into the calculator or miss copyed a number.

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  • $\begingroup$ The derivation is good. What did you set for $r$? $\endgroup$ – npojo Sep 24 '18 at 20:24
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Notwithstanding the fact that you could never orbit within the Earth's atmosphere due to friction (air drag) and the fact that at 1000 ft you would hit mountains, your derivation is sound and a good approximation for very low orbits. Of course as R increased, g diminishes so it would lose accuracy, and energy would not be conserved. For a theoretical 1000 ft above sea level orbit, the result would be .0024% too low. But for a geosynchronous satellite at 35,786 km above sea level, the result would be 61% too low, rendering the formula of little use in the real world.

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  • $\begingroup$ Small typo there. I ment km not ft $\endgroup$ – K Ferreira Sep 26 '18 at 19:11
  • $\begingroup$ @KFerreira At least we're clearing the mountains and atmosphere! :) $\endgroup$ – Stuart Van Horne Sep 26 '18 at 19:20
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The method you suggest does not work because

$$ g = 9.81\,\mbox{m s}^{-2} = \frac{GM_{earth}}{R^2_{earth}}\,. $$

Using $v = \sqrt{g*r}$ for $r > R_{earth}$ will result in the total mechanical energy of the orbiting object not being conserved. This is not allowed since gravity is a conservative force, i.e. an object moving under the influence of the gravitational force has its total mechanical energy conserved.

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  • $\begingroup$ Now I see my mistake. I needed to calculate the acceleration due to gravity for $r>R_Earth$. $\endgroup$ – K Ferreira Sep 27 '18 at 15:10
  • $\begingroup$ Glad to see you know where the error in your thinking occurred. $\endgroup$ – Physics_Et_Al Sep 27 '18 at 17:12

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