A satellite's escape velocity is 6.mi./sec, the radius of the earth is 3960 mi, and the earth's gravitational constant (g) is 32.2 ft./sec^2. How far is the satellite from the surface of the earth?

Answer 1

PElocation+KEescape=0
-GMe/(h+re)+1/2 v^2=0
but GMe/re= g and....
-g re/(h+re)=-1/2 v^2

solve for h check my thinking.

Answer 2

Using conservation of energy, the following equation can be written for any satellite (or planet) "caught" in a gravitational field:
(1/2)MVe^2-(GM1M2)/Re=-(GM1M2)/Rmax
where Rmax is the maximum altitude of the satellite and Ve is its escape velocity.
Solving for Vesc gives this useful equation:
Vesc=sqrt((2GMe)/(Re+h))
here h is any additional distance the satellite is from the surface of the Earth.
solving for h:
((2GMe)/Vesc^2)-Re=h
now all you will need to do is convert the terms and solve.

Answer 3

thank you bobpursley, and Anne so much for your help. you are the best