Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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