Question

A satellite orbits earth at 800 m from the earth's center. Gravity at this location is 6.2 m/s^2. What is the velocity of the satellite

Answer 1

Recall that the magnitude of the acceleration $a$ of a particle moving with speed $v$ in a circular path around a point at a distance $R$ away from the particle is given by

$a=\dfrac{v^2}R$

So, the satellite has velocity

$6.2\,\dfrac{\rm m}{\mathrm s^2}=\dfrac{v^2}{800\,\rm m}\implies v=70.4\,\dfrac{\rm m}{\rm s}$

pointing in the direction tangent to the circular path.