Apollo spacecraft Homework Help

[cverret311]

Homework Statement

On April 11, 1970, Apollo 13 was launched at Kennedy Space Flight center, Florida LC 39A. Appollo 13 carried Astronaunts James A. Lovel, John L Swigert, and Fred W. Haise on their voyage to the moon. However, two days into the flight and 321,860 km from Earth when stirring the oxygen tanks, the Teflon-insulated wires that provided electricity to the stirrer motor were damaged, causing a large fire when electricity passed through them. The fire heated the surrounding oxygen, increasing the pressure inside the tank above its 1000 PSI limit, causing the tanks to explode. This explosion caused Apollo 13 to abandon its moon voyage and to try return back to Earth. To return to Earth, the astronaunts aboard Apollo 13 used the moon to slingshot them back to Earth. The spacecraft had a total mass 68,000 kg (including the 1715.2 kg of fuel and the 3 astronaunts) and was capable of producing 689 kN of thrust. Using this information, determine how long it would take for the Astronaunts traveling at 24,000 mph to perform an emergency burn and return to Earth traveling at 24,000 mph. In other words, how long would it take the Apollo spacecraft to go from 24,000 mph in one direction to 24,000 mph in the other direction. Note: Make time positive.

Homework Equations

(1)Fdt= mvf - mvi

F=Force
dt= change in time
vf=final velocity
vi= initial velocity


(2) Fnet = dp/dt

dp= change in momentum
dt= change in time
Fnet= force

(3) p=mv

p=momentum
m=mass
v=velocity

The Attempt at a Solution

I rearranged the first equation to get dt= mvf-mvi /F

Since the questions asks how long would it take the Apollo spacecraft to go from 24,000 mph in one direction to 24,000 mph in the other direction; I canceled out vf and vi and was left with:
dt= 68000kg/689kN = 98.7s

 

[JaredJames]

I haven't checked through everything yet, but a quick skim through tells me two things:

1) The question gives far too much useless info.

2) Your final maths on the last line of the attempt is wrong. You didn't allow for the k in kN so you are a factor of 1000 out.

 

[JaredJames]

OK, convert mph to m/s for ease so everything is SI.

Now, remember the change in momentum is not zero. The Net momentum is zero.

The change in momentum = p = m(vf - vi).
Note: vf = negative, vi = positive. Therefore in the above equation you get p = m(vf - - vi).

Once you have that, you have the force and you have the change in momentum you can calculate the time.

I could be way off on this one, it's late and I can't get seem to think straight, but if you follow this, you get an answer. Whether or not it's reasonable I don't know. I don't do space flight, but it appears ok if those are the equations you are working with.

Give it a go.

Do you know the actual answer?