Answer
The speed depends on $~~\frac{1}{\sqrt{r}}$
Work Step by Step
We can find an expression for the speed:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
$v^2 = \frac{GM}{r}$
$v = \sqrt{\frac{GM}{r}}$
The speed depends on $~~\frac{1}{\sqrt{r}}$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 13 - Gravitation - Problems - Page 383: 65d
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