Question

With what speed must a satellite be moving in order to maintain a
circular orbit 3360 km above the surface of the Earth?

Answer 1

Answer: $V=10891.39 m/s$

Explanation:

Since we are told the satellite movs in a circular orbit,  we can use the equation of velocity in the case of uniform circular motion:

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

Where:

$V=$ is the velocity of the satellite

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$r=3360 km \frac{1000 m}{1 km}=3360000 m$ is the radius of the orbit

Solving with the given data:

$V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{3360000 m}}$

Finally:

$V=10891.39 m/s$