Question

A satellite moves in a stable circular orbit with speed Vo at a distance R from the center of a planet. For this satellite to move in a stable circular orbit a distance 2R from the center of the planet, the speed of the satellite must be??
I said that F=ma, but m doesn't matter since it's constant. So, a0=a1. a=v^2/r. So V1^2/2R = V0^2/R. I ended up with V1 = V0sqrt(2). But that's not the answer. All the multiple choice answers have sqrt's, 2's, and V0's scattered around, but none are what I have. What did I do wrong??

Answer 1

Answer:

The new speed must be $\frac{V_0}{\sqrt{2}}$

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:

$G \frac{Mm}{R^2}= m\frac{V_0^2}{R}$

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:

$V_0 = \sqrt{\frac{GM}{R}}$

Now, if we want the satellite to orbit at a distance of 2R, the new tangential speed must be:

$V' = \sqrt{\frac{GM}{2R}}=\frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}}= \frac{V_0}{\sqrt{2}}$