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anonymous
 3 years ago
At what distance \(h\) above the surface of the Earth is the freefall 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?
anonymous
 3 years ago
At what distance \(h\) above the surface of the Earth is the freefall 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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The answer I get is \(8.12\times 10^7m\) which is wrong :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You solved for the radial distance from the center of the earth  you have to subtract the earth's radius from that value.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do you get 9.83E6 for a_1? or would I use 9.8 m/s^2?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm surprised that my a_1 doesn't equal 9.8, because that's what the acceleration at sea level is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't know what you mean by that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[a=\frac{6.67\times 10^{11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}=9.83\times 10^6 m/s^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.06370 meters? That's about four miles... :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL! OOOOooops!!! Let's try that again
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