Rockets 'n' Shit

A rocket is fired vertically from the earth's surface. The fuel is completely used up when the rocket reaches an altitude of 40 km, at which time the velocity of the rocket is 6500 ms^-1. What altitude will the rocket reach before being drawn back to earth due to the effect of gravity?

Note: Newton's Law of Gravitation states that the magnitude of the force between two bodies of respective mass m and M is

GmM/d^2

where d = the distance between the centres of the masses
      G = 6.673 * 10^-11 m^3 kg^-1 s^-2

Oh yeah, the earth's mass is approximately 5.976 * 10^24 kg and its radius is 6378 km. Now fucking answer the question.

May I start by saying that I spent a fair bit of time and brain-power on this, and it's pretty late here, so be gentle if I err.

This question requires familiarity with some simple calculus methods and a basic knowledge of some Physics equations - not too hard if you think about it though. Firstly, a labelled diagram which will be referred to down the track:

http://img139.imageshack.us/my.php?image=1ew4.png


Now for some formula derivation:

Now, we are (or should be) aware that

R = -GmMi/x^2

Which can be expressed as

mai = -GmMi/x^2

Some simple rearrangement of this equation gives

a = -GM/x^2  ------> (the m and i cancel in the denominator)

which will be referred to as Equation 1.

Now, at the earth's surface

W = GmM/r^2

but W = mg, so:

mg = GmM/r^2

Again, we can rearrange this slightly to give

gr^2 = GM

which will now be referred to as Equation 2.

Now, substituting Equation 2 into Equation 1, we have

a = -gr^2/x^2.

You should have been taught that acceleration is commonly thought of as the rate of change of velocity, which can be expressed as:

a = d(1/2*v^2)/dx.

So for this question, we have

d(1/2*v^2)/dx = -gr^2/x^2.

Now onto the question itself -- we will need to use integrals. What we need to do is determine at what distance the velocity is zero - at this distance, the rocket will be drawn back to earth due to the effect of gravity (...theoretically).  When v = 0, let x = d, and when v = 6500 ms^-1, let x = 40 km + the radius of the earth = 6418000 m.

Now,

http://img61.imageshack.us/my.php?image=2vo7.png  (integral-y stuffs)

So,

http://img97.imageshack.us/my.php?image=3el7.png (more integral-y)

If you are not sure of how I went from (http://img61.imageshack.us/my.php?image=2vo7.png) to (http://img97.imageshack.us/my.php?image=3el7.png), then you need to tighten up your calc. skills, son!

Equating,

(6500^2/2) – 0 = 9.8 * 6378000^2[(1/6418000) – (1/d)]

(1/6418000) – (1/d) = 5.300 * 10^-8   

1/d = 1.028 * 10^-7

d = 9726000 m

d = 9726 km

So the altitude (above the earth’s surface) reached by the rocket = 9726 – 6378 (remember the rocket was fired from the centre of the earth)

= 3348 km.

Note that the above part (from “Equating,” down) was essentially the only work we had to do in this question – the rest was figuring out just how to do that work and what equations etc. to use for it.

Now, if anything there confuses you greatly then perhaps you should pay more attention in your class; for example, you shouldn’t ask me, “What is R? What are G, m and i? Why did you use those equations? What the fuck are integrals?” etc. – you should already know the answers to these questions. Or you could give up the particular class if you’re finding it too difficult.

Alternatively, if you haven’t covered some of the calculus methods or the physics equations/derivations I dealt with, you probably aren’t ready to tackle problems of this nature and should confer with your teacher. Godspeed.

F = ma = GmM/x^2
a = GM/x^2

v = INTEGRATE(GM/x^2) = -GM/x + C
v(x=40) = 6500 ; solve for C

v(x) = 0 ; solve for x

>>4
lol oops. disregard that i sucks cocks.

>>3
By the way, I read your post and you fail hard. a =/= dv/dx ... a = dv/dt

>>3
err..i don't know anymore...explain the following plox

You should have been taught that acceleration is commonly thought of as the rate of change of velocity, which can be expressed as:

a = d(1/2*v^2)/dx.

>>6


You're an idiot. I never said a = dv/dx - I said a = d(1/2*v^2)/dx. You're right in saying a = dv/dt, but that isn't the only way to express it.

Now, dv/dx = (dv/dt)*(dt/dx) ---------> (chain rule)

= a*[1/(dx/dt)]

= a/v

so a = v(dv/dx).

[d(1/2*v^2)/dv]*(dv/dx) = a

or

d(1/2*v^2)/dx = a.

1: Google newton's equations.
2: Integrate.
3: ???
4: Profit.

Isn't this just a simple conservation of energy problem?

change in Ek = change in Ep

1/2m(v2² - v1²) = mgz

V2 = 0
Everything is constant except for height (z). Solve.

You guys are idiots for helping this asshole for his homework.

>>10
gtfo. we don't need you and you crazy science shit here.