- #1

JJBladester

Gold Member

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## Homework Statement

An 870-lb satellite is placed in a circular orbit 3973 mi above the surface of the earth. At this elevation the acceleration of gravity is 8.03 ft/s

^{2}. Determine the kinetic energy of the satellite, knowing that its orbital speed is 12,500 mi/h.

## Homework Equations

KE=(1/2)mv

^{2}

## The Attempt at a Solution

W=870-lb

g=8.03ft/s

^{2}

h=(3973mi)*(5280ft/1mi)=2.10e

^{7}ft

v=(12,500mi/hr)*(5280ft/1mi)*(1hr/3600s)=1.83e

^{4}ft/s

KE=(1/2)mv

^{2}=(1/2)(W/g)(v

^{2})

KE=(1/2)*(870/8.03)*(1.83e

^{4})

^{2}=1.8e

^{10}ft-lb

The book's answer is 4.54e

^{9}ft-lb.

Where did I go wrong in calculating the satellite's KE?