Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

PhoenixFire

  • 2 years ago

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.

  • This Question is Closed
  1. whpalmer4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. Diwakar
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you are using 'r' is kilometers ,that's it.

  3. whpalmer4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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\[9.81953 \text{ N}\text {/kg}*(1 \text{ kg m s}^{-2}/1 \text{ N}) = 9.81953 \text { m}/\text{s}^{2}\]

  4. PhoenixFire
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh wow. That was amazingly smart of me. Thank you both!

  5. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy