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=\frac{v^2}R`

So, the satellite has velocity

`6.2\,(\rm m)/(\mathrm s^2)=(v^2)/(800\,\rm m)\implies v=70.4\,(\rm m)/(\rm s)`

pointing in the direction tangent to the circular path.