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JenniferSmart1
Group Title
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?
 one year ago
 one year ago
JenniferSmart1 Group Title
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?
 one year ago
 one year ago

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JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
The answer I get is \(8.12\times 10^7m\) which is wrong :(
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@UnkleRhaukus
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@Jemurray3
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.0
You solved for the radial distance from the center of the earth  you have to subtract the earth's radius from that value.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
do you get 9.83E6 for a_1? or would I use 9.8 m/s^2?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I'm surprised that my a_1 doesn't equal 9.8, because that's what the acceleration at sea level is
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.0
I don't know what you mean by that.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Here is how I calculated the acceleration at sea level: \[a=\frac{Gm_E}{r^2}\] \[a=\frac{6.67\times 10^{11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
\[a=\frac{6.67\times 10^{11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}=9.83\times 10^6 m/s^2\]
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.0
6370 meters? That's about four miles... :)
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
LOL! OOOOooops!!! Let's try that again
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
That worked, Thanks!
 one year ago
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