Oh, sorry, and this question (The answer is A) ![Grin ;D](https://www.atarnotes.com/forum/Smileys/default/grin.gif)
Thank you!!!
And
this one we can answer with a bit of formula. The centripetal acceleration:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?ma=\frac{mv^2}{r}\\a=\frac{v^2}{r})
However, we can obtain an expression for the velocity:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?v^2=\left(\frac{2\pi R}{T}\right)^2=\frac{4\pi^2R^2}{T^2})
Combine THAT with a rearranged version of Kepler's Law:
Edit by Rui: Typo spotted![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{r^3}{T^2}=\frac{GM}{4\pi^2}\implies\frac{4\pi^2R^2}{T^2}=\frac{GM}{R})
So that yields \(v=\frac{GM}{R}\), and we can substitute that back:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?a=\frac{v^2}{r}=\frac{GM}{R^2} )
We can substitute our varying values of R (remember that the radius of the earth must be taken into account) to obtain our numerical answers. Or, we can just notice that a higher orbital radius will lower our centripetal acceleration, and so A must be true by default (indeed, this second approach is more time efficient, and may not even require all the working above)
![Grin ;D](https://www.atarnotes.com/forum/Smileys/default/grin.gif)