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JenniferSmart1

  • 2 years ago

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?

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  1. JenniferSmart1
    • 2 years ago
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    The answer I get is \(8.12\times 10^7m\) which is wrong :(

  2. JenniferSmart1
    • 2 years ago
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    @UnkleRhaukus

  3. JenniferSmart1
    • 2 years ago
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    @Jemurray3

  4. Jemurray3
    • 2 years ago
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    You solved for the radial distance from the center of the earth -- you have to subtract the earth's radius from that value.

  5. JenniferSmart1
    • 2 years ago
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    do you get 9.83E6 for a_1? or would I use 9.8 m/s^2?

  6. JenniferSmart1
    • 2 years ago
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    I'm surprised that my a_1 doesn't equal 9.8, because that's what the acceleration at sea level is

  7. Jemurray3
    • 2 years ago
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    I don't know what you mean by that.

  8. JenniferSmart1
    • 2 years ago
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    Here is how I calculated the acceleration at sea level: \[a=\frac{Gm_E}{r^2}\] \[a=\frac{6.67\times 10^{-11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}\]

  9. JenniferSmart1
    • 2 years ago
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    \[a=\frac{6.67\times 10^{-11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}=9.83\times 10^6 m/s^2\]

  10. Jemurray3
    • 2 years ago
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    6370 meters? That's about four miles... :)

  11. JenniferSmart1
    • 2 years ago
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    LOL! OOOOooops!!! Let's try that again

  12. JenniferSmart1
    • 2 years ago
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    That worked, Thanks!

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