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 one year 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?
 one year 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?

This Question is Closed

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

Jemurray3
 one year 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.

JenniferSmart1
 one year 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?

JenniferSmart1
 one year 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

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

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

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.06370 meters? That's about four miles... :)

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

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