Answer:
[tex]V=\sqrt{\frac{2}{3}gR}[/tex]
Explanation:
Using the gravitational force equation
[tex]F=G\frac{mM}{D^{2}}[/tex]
D is the distance from the center of the earth [tex]D=\frac{R}{2}+R=\frac{3R}{2}[/tex]
Now, we can equal this equation to the centripetal force.
[tex]F_{c}=\frac{mV^{2}}{D}[/tex]
V is the speed of the satellite
[tex]\frac{mV^{2}}{D}=G\frac{mM}{D^{2}}[/tex]
[tex]V^{2}=G\frac{M}{D}[/tex]
[tex]V=\sqrt{G\frac{M}{D}}[/tex]
[tex]V=\sqrt{G\frac{M}{(R*3/2)}}[/tex]
[tex]V=\sqrt{2G\frac{M}{3R}}[/tex]
But if we evaluate the force at the surface of the earth we have:
[tex]F=G\frac{mM}{R^{2}}[/tex]
[tex]mg=G\frac{mM}{R^{2}}[/tex]
[tex]g=G\frac{M}{R^{2}}[/tex]
[tex]gR=G\frac{M}{R}[/tex]
Then we can use this expression into the equation of V
[tex]V=\sqrt{\frac{2}{3}gR}[/tex]
I hope it helps you!
