anonymous
  • anonymous
At what distance \(h\) above the surface of the Earth is the free-fall acceleration half its value at sea level? Here is what I did: \[a=\frac{F_g}{m}=\frac{Gm_E}{r^2}\] therefore.... \[r_2=\sqrt{\frac{Gm}{a/2}}\] where \(r_2\) is the \(h\) that we're looking for. What did I do wrong?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The answer I get is \(8.12\times 10^7m\) which is wrong :(
anonymous
  • anonymous
@UnkleRhaukus
anonymous
  • anonymous
@Jemurray3

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anonymous
  • anonymous
You solved for the radial distance from the center of the earth -- you have to subtract the earth's radius from that value.
anonymous
  • anonymous
do you get 9.83E6 for a_1? or would I use 9.8 m/s^2?
anonymous
  • anonymous
I'm surprised that my a_1 doesn't equal 9.8, because that's what the acceleration at sea level is
anonymous
  • anonymous
I don't know what you mean by that.
anonymous
  • anonymous
Here is how I calculated the acceleration at sea level: \[a=\frac{Gm_E}{r^2}\] \[a=\frac{6.67\times 10^{-11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}\]
anonymous
  • anonymous
\[a=\frac{6.67\times 10^{-11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}=9.83\times 10^6 m/s^2\]
anonymous
  • anonymous
6370 meters? That's about four miles... :)
anonymous
  • anonymous
LOL! OOOOooops!!! Let's try that again
anonymous
  • anonymous
That worked, Thanks!

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