Answer:
The new speed must be [tex]\frac{V_0}{\sqrt{2}}[/tex]
Explanation:
In order for the satellite to be on a stable orbit around the planet, the gravitational attraction must be equal to the centripetal force that keeps the satellite in circular motion:
[tex]G \frac{Mm}{R^2}= m\frac{V_0^2}{R}[/tex]
where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, V0 the speed of the satellite at distance R from the center of the planet.
We can re-write V0, the initial satellite speed, by re-arranging the equation:
[tex]V_0 = \sqrt{\frac{GM}{R}}[/tex]
Now, if we want the satellite to orbit at a distance of 2R, the new tangential speed must be:
[tex]V' = \sqrt{\frac{GM}{2R}}=\frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}}= \frac{V_0}{\sqrt{2}}[/tex]