Answer
a) $v=\sqrt{\frac{GM_E}{r}}$
b) $E=-K.E$
c) Yes.
Work Step by Step
(a) We can find the required speed as
$\frac{mv^2}{r}=\frac{GmM_E}{r^2}$
This simplifies to:
$v=\sqrt{\frac{GM_E}{r}}$
(b) We know that
$E=K.E+U$
$E=\frac{1}{2}mv^2-\frac{GM_Em}{r}$
$\implies E=\frac{1}{2}mv^2-(v^2)m$
$E=-\frac{1}{2}mv^2$
$E=-K.E$
(c) We know that the result of part(b) applies to an object orbiting the Sun. We would change the mass of the Earth to the mass of the Sun, but regardless this mass will be eliminated from the final result.
