Question

A satellite moves at a constant speed in a circular orbit about the center of the

Answer 1

Answer:

$V=\sqrt{\frac{2}{3}gR}$

Explanation:

Using the gravitational force equation

$F=G\frac{mM}{D^{2}}$

D is the distance from the center of the earth $D=\frac{R}{2}+R=\frac{3R}{2}$

Now, we can equal this equation to the centripetal force.

$F_{c}=\frac{mV^{2}}{D}$

V is the speed of the satellite

$\frac{mV^{2}}{D}=G\frac{mM}{D^{2}}$

$V^{2}=G\frac{M}{D}$

$V=\sqrt{G\frac{M}{D}}$

$V=\sqrt{G\frac{M}{(R*3/2)}}$

$V=\sqrt{2G\frac{M}{3R}}$

But if we evaluate the force at the surface of the earth we have:

$F=G\frac{mM}{R^{2}}$

$mg=G\frac{mM}{R^{2}}$

$g=G\frac{M}{R^{2}}$

$gR=G\frac{M}{R}$

Then we can use this expression into the equation of V

$V=\sqrt{\frac{2}{3}gR}$

I hope it helps you!