A satellite in a circular orbit experiences a centripetal acceleration of #8.62# #m##/s^2#. The tangential speed of the satellite is #7.65 * 10^3 m##/s#. What is the altitude of the satellite?

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Answer 1 · 2018-06-05T16:15:20

The altitude is #=419km#

The centripetal acceleration is #a_N=8.62ms^-2#

The tangential speed is #v=7.65*10^3ms^-1#

Apply the equation

#a_N=v^2/r#

The radius (distance between the satellite and the center of the earth) is

#r=v^2/a_N=(7.65*10^3)^2/8.62=6.789*10^6m#

The radius of the earth is

#R=6370*10^3m=6.37*10^6m#

The altitude of the satellite is

#h=r-R=(6.789-6.37)*10^6m=0.419*10^6m=419km#