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?

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.

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.

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

To find the distance of the satellite from the surface of the earth, we need to first convert the satellite's escape velocity from miles per second to feet per second. Then we can use the equation for escape velocity to solve for the distance.

1. Convert the escape velocity from miles per second to feet per second:
Since 1 mile = 5280 feet and 1 second = 1 second, we can multiply the escape velocity by both conversion factors:
6 mi./sec * 5280 ft./mi = 31,680 ft./sec.

2. Use the equation for escape velocity:
Escape velocity = √(2 * g * r),
where g is the gravitational constant and r is the radius of the Earth.

Plugging in the values we have:
31,680 ft./sec = √(2 * 32.2 ft./sec^2 * 3960 mi * 5280 ft./mi).

3. Simplify the equation:
31,680 ft./sec = √(2 * 32.2 ft./sec^2 * 3960 mi * 5280 ft./mi).
31,680 ft./sec = √(2 * 32.2 ft./sec^2 * 3960 * 5280 mi * ft./mi).
31,680 ft./sec = √(2 * 32.2 ft./sec^2 * 3960 * 5280).
31,680 ft./sec = √(2 * 32.2 * 3960 * 5280) ft./sec.
31,680 ft./sec ≈ 37,447.314 ft./sec.

4. Now we can solve for the distance. Rearrange the equation for escape velocity:
Escape velocity = √(2 * g * r).
r = Escape velocity^2 / (2 * g).

Plugging in the known values:
r ≈ (37,447.314 ft./sec)^2 / (2 * 32.2 ft./sec^2).
r ≈ 1,404,672,398.64 ft.^2/sec^2 / 64.4 ft.^2/sec^2.
r ≈ 21,823,220.19 ft.

Therefore, the distance of the satellite from the surface of the Earth is approximately 21,823,220.19 feet.