Fitzpatrick ex. 25b question.. (1 Viewer)

[govich]

hi guys, just checking if anyone's figured out question 16 part c in this exercise .. i've got the first 2 parts, and i know how to get the answer for part c, but i cant figure out why that's the answer

following on from part b, x = 2gR / 2gR - V.. i'm fine with that.
the answer says that the rocket should escape gravitational attraction when 2gR - V < 0 .. ie V > 11200m/s. anyone have a decent explanation for this? this equation for x is how high the rocket will rise before coming back, so all i can think of is that if the denominator is negative or 0, it'll make x erroneous (negative distance or infinite).
buuuut if you work with this like i did, the answer should be v is greater than OR equal to 11200 (as the denominator is less than OR equal to 0). anyone?

 

[govich]

OK incase some of u dont have Fitzpatrick or whatever, here's the question:

"A rocket is fired vertically from the earth's surface with an initial speed V. Assuming negligible air resistance (a) find an expression for v(x) which gives the velocity at a distance x from the center of the earth, (b) find how high the rocket will rise, (c) find the magnitude of V in order that the rocket should escape from the gravitational attraction and never return. Evaluate this particular value of V, given that the radius R of the earth is 6400km.
(Since the acceleration due to gravity is inversely proportional to the square of the distance from O, its magnitude at any point P is equal to k/x)"

 

[Zarathustra]

Could you post up the worked solutions fo parts a) and b)

 

[govich]

ahhh i've figured this one out since i posted.. i can still post up the answers if anyone else is having trouble though.

 

[CrashOveride]

yeah post em up... i was thinking u might need physics know how to do this ?