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PhoenixFire Group Title

Confused here... \[F=mg=m \frac{GM}{r^2}\] We know 'g' is an acceleration which is \(ms^{-2}\) Now to calculate 'g' we use \[g=\frac{GM}{r^2}\] \(G=6.674 \times 10^{-11}\) \(M=5.972 \times 10^{24}\) \(r=6371\) putting all this into that equation I get \(9.81 \times 10^6\) Acceleration due to gravity is definitely not 9 million. Can anyone tell where I messed up? or what I'm not understanding.

  • one year ago
  • one year ago

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  1. whpalmer4 Group Title
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    well, you don't have any units on any of this, which is problematic.

    • one year ago
  2. Diwakar Group Title
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    you are using 'r' is kilometers ,that's it.

    • one year ago
  3. whpalmer4 Group Title
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    If you'd kept the units on the numbers while working the problem, you would have discovered the issue yourself. \[g = \frac{GM}{r^2} = \frac{(6.674*10^{-11} \text{ N m}^2\text{/kg} ^2)(5.972*10^{24}\text { kg})}{(6371 \text{ km})^2 } \rightarrow [num] \frac{\text{N kg }\color{red}{\text {m}^2} }{\text{ kg}^2\color{red}{\text{ km}^2 }} \]and you see right away that a conversion hasn't been applied because the units don't cancel/combine properly. Include the missing conversion factor and it comes out properly, as we see below: \[g = \frac{GM}{r^2} = \frac{(6.674*10^{-11} \text{ N m}^2\text{/kg} ^2)(5.972*10^{24}\text { kg})}{((6371 \text{ km})(1000\text{ m}/1 \text{ km}))^2 }\]\[ = 9.81953 \text{ N}\text {/kg}\]or if you prefer\[9.81953 \text{ N}\text {/kg}*(1 \text{ kg m s}^{-2}/1 \text{ N}) = 9.81953 \text { m}/\text{s}^{2}\]

    • one year ago
  4. PhoenixFire Group Title
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    Oh wow. That was amazingly smart of me. Thank you both!

    • one year ago
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